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Chapter 8 Resource Masters
Copyright © by The McGraw-Hill Companies, Inc. All rights reserved. Permission is granted to reproduce the material contained herein on the condition that such materials be reproduced only for classroom use; be provided to students, teachers, and families without charge; and be used solely in conjunction with the Glencoe Geometry program. Any other reproduction, for sale or other use, is expressly prohibited.
Send all inquiries to:Glencoe/McGraw-Hill8787 Orion PlaceColumbus, OH 43240 - 4027
ISBN: 978-0-07-890517-9
MHID: 0-07-890517-6
Printed in the United States of America.
4 5 6 7 8 9 10 11 12 REL 19 18 17 16 15 14 13 12 11
CONSUMABLE WORKBOOKS Many of the worksheets contained in the Chapter Resource Masters
booklets are available as consumable workbooks in both English and Spanish.
ISBN10 ISBN13
Study Guide and Intervention Workbook 0-07-890848-5 978-0-07-890848-4
Homework Practice Workbook 0-07-890849-3 978-0-07-890849-1
Spanish Version
Homework Practice Workbook 0-07-890853-1 978-0-07-890853-8
ANSWERS FOR WORKBOOKS The answers for Chapter 8 of these workbooks can be found in the back of this Chapter Resource Masters booklet.
StudentWorks PlusTM This CD-ROM includes the entire Student Edition text along with the English workbooks listed above.
TeacherWorks PlusTM All of the materials found in this booklet are included for viewing, printing,
and editing in this CD-ROM.
Spanish Assessment Masters (ISBN10: 0-07-890856-6, ISBN13: 978-0-07-890856-9) These masters contain a Spanish version of Chapter 8 Test Form 2A and Form 2C.
iii
Contents
Teacher’s Guide to Using the Chapter 8
Resource Masters .........................................iv
Chapter Resources Chapter 8 Student-Built Glossary ...................... 1
Chapter 8 Anticipation Guide (English) ............. 3
Chapter 8 Anticipation Guide (Spanish) ............ 4
Lesson 8-1Geometric Mean
Study Guide and Intervention ............................ 5
Skills Practice .................................................... 7
Practice .............................................................. 8
Word Problem Practice ..................................... 9
Enrichment ...................................................... 10
Lesson 8-2The Pythagorean Theorem and
Its Converse
Study Guide and Intervention .......................... 11
Skills Practice .................................................. 13
Practice .......................................................... 14
Word Problem Practice ................................... 15
Enrichment ...................................................... 16
Spreadsheet Activity ........................................ 17
Lesson 8-3Special Right Triangles
Study Guide and Intervention .......................... 18
Skills Practice .................................................. 20
Practice .......................................................... 21
Word Problem Practice ................................... 22
Enrichment ...................................................... 23
Lesson 8-4Trigonometry
Study Guide and Intervention .......................... 24
Skills Practice .................................................. 26
Practice .......................................................... 27
Word Problem Practice ................................... 28
Enrichment ...................................................... 29
Lesson 8-5Angles of Elevation and Depression
Study Guide and Intervention .......................... 30
Skills Practice .................................................. 32
Practice .......................................................... 33
Word Problem Practice ................................... 34
Enrichment ...................................................... 35
Lesson 8-6The Law of Sines and Law of Cosines
Study Guide and Intervention .......................... 36
Skills Practice .................................................. 38
Practice .......................................................... 39
Word Problem Practice ................................... 40
Enrichment ...................................................... 41
Graphing Calculator Activity ............................ 42
Lesson 8-7Vectors
Study Guide and Intervention .......................... 43
Skills Practice .................................................. 45
Practice .......................................................... 46
Word Problem Practice ................................... 47
Enrichment ...................................................... 48
AssessmentStudent Recording Sheet ................................ 49
Rubric for Extended-Response ....................... 50
Chapter 8 Quizzes 1 and 2 ............................. 51
Chapter 8 Quizzes 3 and 4 ............................. 52
Chapter 8 Mid-Chapter Test ............................ 53
Chapter 8 Vocabulary Test ............................. 54
Chapter 8 Test, Form 1 ................................... 55
Chapter 8 Test, Form 2A ................................. 57
Chapter 8 Test, Form 2B ................................. 59
Chapter 8 Test, Form 2C ................................ 61
Chapter 8 Test, Form 2D ................................ 63
Chapter 8 Test, Form 3 ................................... 65
Chapter 8 Extended-Response Test ............... 67
Standardized Test Practice ............................. 68
Answers ........................................... A1–A36
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Teacher’s Guide to Using the
Chapter 8 Resource MastersThe Chapter 8 Resource Masters includes the core materials needed for Chapter 8. These materials include worksheets, extensions, and assessment options. The answers for these pages appear at the back of this booklet.
All of the materials found in this booklet are included for viewing and printing on the
TeacherWorks PlusTM CD-ROM.
Chapter Resources
Student-Built Glossary (pages 1–2) These
masters are a student study tool that
presents up to twenty of the key vocabulary
terms from the chapter. Students are to
record definitions and/or examples for each
term. You may suggest that students
highlight or star the terms with which they
are not familiar. Give this to students before
beginning Lesson 8-1. Encourage them to
add these pages to their mathematics study
notebooks. Remind them to complete the
appropriate words as they study each lesson.
Anticipation Guide (pages 3–4) This master, presented in both English and Spanish, is a survey used before beginning the chapter to pinpoint what students may or may not know about the concepts in the chapter. Students will revisit this survey after they complete the chapter to see if
their perceptions have changed.
Lesson Resources
Study Guide and Intervention These masters provide vocabulary, key concepts, additional worked-out examples and Check Your Progress exercises to use as a reteaching activity. It can also be used in conjunction with the Student Edition as an instructional tool for students who have been absent.
Skills Practice This master focuses more on the computational nature of the lesson. Use as an additional practice option or as homework for second-day teaching of the lesson.
Practice This master closely follows the types of problems found in the Exercises section of the Student Edition and includes word problems. Use as an additional practice option or as homework for second-day teaching of the lesson.
Word Problem Practice This master includes additional practice in solving word problems that apply the concepts of the lesson. Use as an additional practice or as homework for second-day teaching of the lesson.
Enrichment These activities may extend the concepts of the lesson, offer an historical or multicultural look at the concepts, or widen students’ perspectives on the mathematics they are learning. They are written for use with all levels of students.
Graphing Calculator or Spreadsheet
Activities These activities present ways in which technology can be used with the concepts in some lessons of this chapter. Use as an alternative approach to some concepts or as an integral part of your lesson presentation.
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Assessment OptionsThe assessment masters in the Chapter 8
Resource Masters offer a wide range of assessment tools for formative (monitoring) assessment and summative (final) assessment.
Student Recording Sheet This master corresponds with the standardized test practice at the end of the chapter.
Extended–Response Rubric This master provides information for teachers and students on how to assess performance on open-ended questions.
Quizzes Four free-response quizzes offer assessment at appropriate intervals in the chapter.
Mid-Chapter Test This 1-page test provides an option to assess the first half of the chapter. It parallels the timing of the Mid-Chapter Quiz in the Student Edition and includes both multiple-choice and free-response questions.
Vocabulary Test This test is suitable for all students. It includes a list of vocabulary words and 10 questions to assess students’ knowledge of those words. This can also be used in conjunction with one of the leveled chapter tests.
Leveled Chapter Tests
• Form 1 contains multiple-choice questions and is intended for use with below grade level students.
• Forms 2A and 2B contain multiple-choice questions aimed at on grade level stu-dents. These tests are similar in format to offer comparable testing situations.
• Forms 2C and 2D contain free-response questions aimed at on grade level students. These tests are similar in format to offer comparable testing situations.
• Form 3 is a free-response test for use with above grade level students.
All of the above mentioned tests include a free-response Bonus question.
Extended-Response Test Performance assessment tasks are suitable for all students. Sample answers and a scoring rubric are included for evaluation.
Standardized Test Practice These three pages are cumulative in nature. It includes three parts: multiple-choice questions with bubble-in answer format, griddable questions with answer grids, and short-answer free-response questions.
Answers• The answers for the Anticipation Guide
and Lesson Resources are provided as reduced pages.
• Full-size answer keys are provided for the assessment masters.
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Chapter 8 1 Glencoe Geometry
Student-Built Glossary8
Vocabulary TermFound
on PageDefinition/Description/Example
angle of depression
angle of elevation
component form
cosine
geometric mean
Law of Cosines
Law of Sines
magnitude
(continued on the next page)
This is an alphabetical list of the key vocabulary terms you will learn in Chapter 8.
As you study the chapter, complete each term’s definition or description.
Remember to add the page number where you found the term. Add these pages to
your Geometry Study Notebook to review vocabulary at the end of the chapter.
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Chapter 8 2 Glencoe Geometry
Student-Built Glossary (continued)8
Vocabulary TermFound
on PageDefinition/Description/Example
Pythagorean triple
resultant
sine
tangent
trigonometric ratio
trigonometry
vector
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Chapter 8 3 Glencoe Geometry
Before you begin Chapter 8
• Read each statement.
• Decide whether you Agree (A) or Disagree (D) with the statement.
• Write A or D in the first column OR if you are not sure whether you agree or disagree, write NS (Not Sure).
Anticipation Guide
Right Triangles and Trigonometry
After you complete Chapter 8
• Reread each statement and complete the last column by entering an A or a D.
• Did any of your opinions about the statements change from the first column?
• For those statements that you mark with a D, use a piece of paper to write an example of why you disagree.
8
Step 1
STEP 1A, D, or NS
StatementSTEP 2A or D
1. The geometric mean between two numbers is the positive square root of their product.
2. An altitude drawn from the right angle of a right triangle to its hypotenuse separates the triangle into two congruent triangles.
3. In a right triangle, the length of the hypotenuse is equal to the sum of the lengths of the legs.
4. If any triangle has sides with lengths 3, 4, and 5, then that triangle is a right triangle.
5. If the two acute angles of a right triangle are 45°,
then the length of the hypotenuse is √ " 2 times the
length of either leg.
6. In any triangle whose angle measures are 30°, 60°,
and 90°, the hypotenuse is √ " 3 times as long as the shorter leg.
7. The sine ratio of an angle of a right triangle is equal to the length of the adjacent side divided by the length of the hypotenuse.
8. The tangent of an angle of a right triangle whose sides have lengths 3, 4, and 5 will be smaller than the tangent of an angle of a right triangle whose sides have lengths 6, 8, and 10.
9. Trigonometric ratios can be used to solve problems involving angles of elevation and angles of depression.
10. The Law of Sines can only be used in right triangles.
Step 2
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NOMBRE FECHA PERÍODO
Antes de comenzar el Capítulo 8
• Lee cada enunciado.
• Decide si estás de acuerdo (A) o en desacuerdo (D) con el enunciado.
• Escribe A o D en la primera columna O si no estás seguro(a) de la respuesta, escribe NS (No estoy seguro(a).
PASO 1A, D o NS
EnunciadoPASO 2A o D
1. La media geométrica de dos números es la raíz cuadrada positiva de su producto.
2. Una altitud que se dibuja desde el ángulo recto de un triángulo rectángulo hasta su hipotenusa divide el triángulo en dos triángulos congruentes.
3. En un triángulo rectángulo, la longitud de la hipotenusa es igual a la suma de las longitudes de los catetos.
4. Si un triángulo rectángulo tiene lados con longitudes 3, 4 y 5, entonces cualquier triángulo es un triángulo rectángulo.
5. Si los dos ángulos agudos de un triángulo rectángulo son de 45°, entonces la longitud de la hipotenusa es √ " 2 veces la longitud de cualquiera de los catetos.
6. En cualquier triángulo cuyas medidas de ángulos sean de 30°, 60° y 90°, la hipotenusa es √ " 3 veces tan larga como el cateto más corto.
7. La razón del seno para un ángulo en un triángulo rectángulo es igual a la longitud del lado adyacente dividido entre la longitud de la hipotenusa.
8. La tangente para un ángulo de un triángulo rectángulo cuyos lados tienen longitudes 3, 4 y 5 será menor que la tangente de un ángulo en un triángulo rectángulo cuyos lados tienen longitudes 6, 8 y 10.
9. Las razones trigonométricas se pueden usar para resolver problemas de ángulos de elevación y ángulos de depresión.
10. La ley de los senos sólo se puede usar con triángulos rectángulos.
Después de completar el Capítulo 8
• Vuelve a leer cada enunciado y completa la última columna con una A o una D.
• ¿Cambió cualquiera de tus opiniones sobre los enunciados de la primera columna?
• En una hoja de papel aparte, escribe un ejemplo de por qué estás en desacuerdo con los enunciados que marcaste con una D.
Ejercicios preparatorios
Triángulos rectángulos y trigonometría
Paso 1
Paso 2
8
Capítulo 8 4 Geometría de Glencoe
Lesso
n 8
-1
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Chapter 8 5 Glencoe Geometry
Study Guide and Intervention
Geometric Mean
Geometric Mean The geometric mean between two numbers is the positive square root of their product. For two positive numbers a and b, the geometric mean of a and b is
the positive number x in the proportion a −
x =
x −
b . Cross multiplying gives x2 = ab, so x = √ ## ab .
Find the geometric mean between each pair of numbers.
a. 12 and 3
x = √ ## ab Definition of geometric mean
= √ ### 12 . 3 a = 12 and b = 3
= √ ##### (2 . 2 . 3) . 3 Factor.
= 6 Simplify.
The geometric mean between 12 and 3 is 6.
b. 8 and 4
x = √ ## ab Definition of geometric mean
= √ ## 8 . 4 a = 8 and b = 4
= √ #### (2 . 4) . 4 Factor.
= √ ### 16 . 2 Associative Property
= 4 √ # 2 Simplify.
The geometric mean between 8 and 4
is 4 √ # 2 or about 5.7.
Exercises
Find the geometric mean between each pair of numbers.
1. 4 and 4 2. 4 and 6
3. 6 and 9 4. 1 − 2 and 2
5. 12 and 20 6. 4 and 25
7. 16 and 30 8. 10 and 100
9. 1 − 2 and 1 −
4 10. 17 and 3
11. 4 and 16 12. 3 and 24
8-1
Example
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Chapter 8 6 Glencoe Geometry
z
y
x
15
R
Q P
S
25
Study Guide and Intervention (continued)
Geometric Mean
Geometric Means in Right Triangles In the diagram, △ ABC ∼ △ ADB ∼ △ BDC. An altitude to the hypotenuse of a right triangle forms two right triangles. The two triangles are similar and each is similar to the original triangle.
Use right △ ABC with
−−
BD ⊥ −−
AC . Describe two geometric
means.
a. △ ADB ∼ △ BDC so AD −
BD = BD
− CD
.
In △ABC, the altitude is the geometric
mean between the two segments of the
hypotenuse.
b. △ ABC ∼ △ ADB and △ ABC ∼ △ BDC,
so AC
− AB
= AB −
AD and
AC −
BC =
BC −
DC .
In △ ABC, each leg is the geometric
mean between the hypotenuse and the
segment of the hypotenuse adjacent to
that leg.
Find x, y, and z.
15 = √ &&&& RP ' SP Geometric Mean (Leg) Theorem
15 = √ && 25x RP = 25 and SP = x
225 = 25x Square each side.
9 = x Divide each side by 25.
Then
y = RP – SP
= 25 – 9
= 16
z = √ &&&& RS ' RP Geometric Mean (Leg) Theorem
= √ &&& 16 ' 25 RS = 16 and RP = 25
= √ && 400 Multiply.
= 20 Simplify.
Exercises
Find x, y, and z to the nearest tenth.
1. x
1 3
2.
z
xy 5
2 3. zx
y
81
4.
xy
1
√3
√"12 5.
x
z y
2
2
6. x z
y
62
8-1
CD
B
A
Example 1 Example 2
Lesso
n 8
-1
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Chapter 8 7 Glencoe Geometry
Skills Practice
Geometric Mean
Find the geometric mean between each pair of numbers.
1. 2 and 8 2. 9 and 36 3. 4 and 7
4. 5 and 10 5. 28 and 14 6. 7 and 36
Write a similarity statement identifying the three similar triangles in the figure.
7.
C
D
B
A 8.
L
M
N
P
9.
G
E H
F
10.
R T
S
U
Find x, y and z.
11.
3 9
yx
z
12.
10
4
y
x
z
13.
15
4
y
xz
14.
2
5y
x
z
8-1
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Chapter 8 8 Glencoe Geometry
Practice
Geometric Mean
Find the geometric mean between each pair of numbers.
1. 8 and 12 2. 3 and 15 3. 4 − 5 and 2
Write a similarity statement identifying the three similar triangles in the figure.
4. U
TA
V
5.
KL
J M
Find x, y, and z.
6.
23
z
xy
8 7.
zx
y
625
8.
x
y
2
3
z
9.
x y
10z
20
10. CIVIL An airport, a factory, and a shopping center are at the vertices of a right triangle
formed by three highways. The airport and factory are 6.0 miles apart. Their distances
from the shopping center are 3.6 miles and 4.8 miles, respectively. A service road will be
constructed from the shopping center to the highway that connects the airport and
factory. What is the shortest possible length for the service road? Round to the nearest
hundredth.
8-1
Lesso
n 8
-1
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Chapter 8 9 Glencoe Geometry
1. SQUARES Wilma has a rectangle of
dimensions ℓ by w. She would like to
replace it with a square that has the
same area. What is the side length of
the square with the same area as
Wilma’s rectangle?
2. EQUALITY Gretchen computed the
geometric mean of two numbers. One of
the numbers was 7 and the geometric
mean turned out to be 7 as well. What
was the other number?
3. VIEWING ANGLE A photographer
wants to take a picture of a beach front.
His camera has a viewing angle of 90°
and he wants to make sure two palm
trees located at points A and B in the
figure are just inside the edges of the
photograph.
Wal
kway
90 ft 40 ftA B
x
A B
x
He walks out on a walkway that goes
over the ocean to get the shot. If his
camera has a viewing angle of 90°, at
what distance down the walkway should
he stop to take his photograph?
4. EXHIBITIONS A museum has a famous
statue on display. The curator places the
statue in the corner of a rectangular
room and builds a 15-foot-long railing in
front of the statue. Use the information
below to find how close visitors will be
able to get to the statue.
Statue12 ft
15 ft9 ft
x
5. CLIFFS A bridge connects to a tunnel
as shown in the figure. The bridge is
180 feet above the ground. At a distance
of 235 feet along the bridge out of the
tunnel, the angle to the base and
summit of the cliff is a right angle.
Cliff
235 ft
180 f
t
dx
h
a. What is the height of the cliff? Round
to the nearest whole number.
b. How high is the cliff from base to
summit? Round to the nearest whole
number.
c. What is the value of d? Round to the
nearest whole number.
Word Problem Practice
Geometric Mean
8-1
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Chapter 8 10 Glencoe Geometry
Mathematics and Music
Pythagoras, a Greek philosopher who lived during the sixth century B.C., believed that all nature, beauty, and harmony could be expressed by whole-number relationships. Most people remember Pythagoras for his teachings about right triangles. (The sum of the squares of the legs equals the square of the hypotenuse.) But Pythagoras also discovered relationships between the musical notes of a scale. These relationships can be expressed as ratios.
C D E F G A B C′1 −
1
8 −
9 4
− 5
3 −
4 2
− 3
3 −
5
8 −
15 1
− 2
When you play a stringed instrument, The C string can be usedyou produce different notes by placing to produce F by placing
your finger on different places on a string. a finger 3 −
4 of the way
This is the result of changing the length along the string.of the vibrating part of the string.
Suppose a C string has a length of 16 inches. Write and solve
proportions to determine what length of string would have to
vibrate to produce the remaining notes of the scale.
1. D 2. E 3. F
4. G 5. A 6. B
7. C′
8. Complete to show the distance between finger positions on the 16-inch
C string for each note. For example, C(16) - D (14 2 − 9 ) = 1
7 −
9 .
C D E F G A B C′
34
of C string
Enrichment
1 7−9
in.
8-1
Less
on
8-2
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Chapter 8 11 Glencoe Geometry
Study Guide and Intervention
The Pythagorean Theorem and Its Converse
The Pythagorean Theorem In a right triangle, the sum of the squares of the lengths of the legs equals the square of the length of
the hypotenuse. If the three whole numbers a, b, and c satisfy the equation
a2 + b2 = c2, then the numbers a, b, and c form a
Pythagorean triple.
ca
bA C
B
a. Find a.
a
12
13
AC
B
a2 + b2 = c2 Pythagorean Theorem
a2 + 122 = 132 b = 12, c = 13
a2 + 144 = 169 Simplify.
a2 = 25 Subtract.
a = 5 Take the positive square root
of each side.
b. Find c.
c
30
20
AC
B
a2 + b2 = c2 Pythagorean Theorem
202 + 302 = c2 a = 20, b = 30
400 + 900 = c2 Simplify.
1300 = c2 Add.
√ "" 1300 = c Take the positive square root
of each side.
36.1 ≈ c Use a calculator.
Find x.
1.
x
3 3 2.
x
159
3.
x
6525
4. x
5
9
4
9 5.
x
33
16
6. x
1128
Use a Pythagorean Triple to find x.
7. 8
17
x
8. 45
24x
9. 28
96
x
8-2
Example
Exercises
△ ABC is a right triangle.
so a2 + b2 = c2.
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Chapter 8 12 Glencoe Geometry
c
ab
A
C
B
Study Guide and Intervention (continued)
The Pythagorean Theorem and Its Converse
Converse of the Pythagorean Theorem If the sum of the squares of the lengths of the two shorter sides of a triangle equals the square of the lengths of the longest side, then the triangle is a right triangle.
You can also use the lengths of sides to classify a triangle. If a2 + b2 = c2, then
if a2 + b2 = c2 then △ABC is a right triangle. △ABC is a right triangle.
if a2 + b2 > c2 then △ABC is acute. if a2 + b2 < c2 then △ABC is obtuse.
Determine whether △PQR is a right triangle.
a2 + b2 " c2 Compare c2 and a2 + b2
102 + (10 √ $ 3 )2 " 202 a = 10, b = 10 √ $ 3 , c = 20
100 + 300 " 400 Simplify.
400 = 400% Add.
Since c2 = and a2 + b2, the triangle is a right triangle.
Exercises
Determine whether each set of measures can be the measures of the sides of a
triangle. If so, classify the triangle as acute, obtuse, or right. Justify your answer.
1. 30, 40, 50 2. 20, 30, 40 3. 18, 24, 30
4. 6, 8, 9 5. 6, 12, 18 6. 10, 15, 20
7. √ $ 5 , √ $$ 12 , √ $$ 13 8. 2, √ $ 8 , √ $$ 12 9. 9, 40, 41
2010
10√3
8-2
Example
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Lesso
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-2
Chapter 8 13 Glencoe Geometry
Skills Practice
The Pythagorean Theorem and Its Converse
Find x.
1. x
12
9
2. x
12
13 3.
x
12
32
4. x12.5
25
5.
x
9 9
8
6.
x
31
14
Use a Pythagorean Triple to find x.
7.
5
12
x
8.
8
10
x 9.
20
12
x
10.
65
25
x
11. 12.
50
48
x
Determine whether each set of numbers can be measure of the sides of a triangle.
If so, classify the triangle as acute, obtuse, or right. Justify your answer.
13. 7, 24, 25 14. 8, 14, 20 15. 12.5, 13, 26
16. 3 √ " 2 , √ " 7 , 4 17. 20, 21, 29 18. 32, 35, 70
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Chapter 8 14 Glencoe Geometry
Practice
The Pythagorean Theorem and Its Converse
Find x.
1. x
13
23
2.
x
3421
3. x
26
2618
4.
x
34
22
5. x
16
14
6. x
2424
42
Use a Pythagorean Triple to find x.
7.
36
27
x
8.
120
136
x
9.
65
39
x
10. 42
150
x
Determine whether each set of numbers can be measure of the sides of a triangle.
If so, classify the triangle as acute, obtuse, or right. Justify your answer.
11. 10, 11, 20 12. 12, 14, 49 13. 5 √ " 2 , 10, 11
14. 21.5, 24, 55.5 15. 30, 40, 50 16. 65, 72, 97
17. CONSTRUCTION The bottom end of a ramp at a warehouse is
10 feet from the base of the main dock and is 11 feet long. How
high is the dock?
11 ft?
dock
ramp
10 ft
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Lesso
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-2
Chapter 8 15 Glencoe Geometry
Word Problem Practice
The Pythagorean Theorem and Its Converse
1. SIDEWALKS Construction workers are building a marble sidewalk around a park that is shaped like a right triangle. Each marble slab adds 2 feet to the length of the sidewalk. The workers find that exactly 1071 and 1840 slabs are required to make the sidewalks along the short sides of the park. How many slabs are required to make the sidewalk that runs along the long side of the park?
2. RIGHT ANGLES Clyde makes a triangle using three sticks of lengths 20 inches, 21 inches, and 28 inches. Is the triangle a right triangle? Explain.
3. TETHERS To help support a flag pole, a 50-foot-long tether is tied to the pole at a point 40 feet above the ground. The tether is pulled taut and tied to an anchor in the ground. How far away from the base of the pole is the anchor?
4. FLIGHT An airplane lands at an airport 60 miles east and 25 miles north of where it took off.
60 mi
25 mi
How far apart are the two airports?
5. PYTHAGOREAN TRIPLES Ms. Jones assigned her fifth-period geometry class the following problem.
Let m and n be two positive integers with m > n. Let a = m2 – n2, b = 2mn, and c = m2 + n2.
a. Show that there is a right triangle with side lengths a, b, and c.
b. Complete the following table.
m n a b c
2 1 3 4 5
3 1
3 2
4 1
4 2
4 3
5 1
c. Find a Pythagorean triple that corresponds to a right triangle with a hypotenuse 252 = 625 units long. (Hint: Use the table you completed for Exercise b to find two positive integers m and n with m > n and m2 + n2 = 625.)
8-2
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Chapter 8 16 Glencoe Geometry
Enrichment
Converse of a Right Triangle TheoremYou have learned that the measure of the altitude from the vertex of
the right angle of a right triangle to its hypotenuse is the geometric
mean between the measures of the two segments of the hypotenuse.
Is the converse of this theorem true? In order to find out, it will help
to rewrite the original theorem in if-then form as follows.
If △ABQ is a right triangle with right angle at Q, then
QP is the geometric mean between AP and PB, where P
is between A and B and −−−
QP is perpendicular to −−
AB .
1. Write the converse of the if-then form of the theorem.
2. Is the converse of the original theorem true? Refer
to the figure at the right to explain your answer.
You may find it interesting to examine the other theorems in
Chapter 8 to see whether their converses are true or false. You will
need to restate the theorems carefully in order to write their
converses.
Q
BP
A
Q
BP
A
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Lesso
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-2
Chapter 8 17 Glencoe Geometry
Spreadsheet Activity
Pythagorean Triples
You can use a spreadsheet to determine whether three whole numbers form a Pythagorean triple.
Use a spreadsheet to determine whether the numbers 12, 16, and 20
form a Pythagorean triple.
Step 1 In cell A1, enter 12. In cell B1, enter 16 and in cell C1, enter 20. The longest side should be entered in column C.
Step 2 In cell D1, enter an equals sign followed by IF(A1^2+B1^2=C1^2,“YES”,“NO”). This will return “YES” if the set of numbers is a Pythagorean triple and will return “NO” if itis not.
The numbers 12, 16, and 20 form a Pythagorean triple.
Use a spreadsheet to determine whether the numbers 3, 6, and 12
form a Pythagorean triple.
Step 1 In cell A2, enter 3, in cell B2, enter 6, and in cell C2, enter 12.
Step 2 Click on the bottom right corner of cell D1 and drag it to D2. This will determine whether or not the set of numbers is a Pythagorean triple.
The numbers 3, 6, and 12 do not form a Pythagorean triple.
Exercises
Use a spreadsheet to determine whether each set of numbers
forms a Pythagorean triple.
1. 14, 48, 50 2. 16, 30, 34 3. 5, 5, 9
4. 4, 5, 7 5. 18, 24, 30 6. 10, 24, 26
7. 25, 60, 65 8. 2, 4, 5 9. 19, 21, 22
10. 18, 80, 82 11. 5, 12, 13 12. 20, 48, 52
8-2
Example 1
Example 2
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Chapter 8 18 Glencoe Geometry
Study Guide and Intervention
Special Right Triangles
Properties of 45°-45°-90° Triangles The sides of a 45°-45°-90° right triangle have a special relationship.
If the leg of a 45°-45°-90°
right triangle is x units, show
that the hypotenuse is x √ " 2 units.
x √
x
x245°
45°
Using the Pythagorean Theorem with
a = b = x, then
c2 = a2 + b2
c2 = x2 + x2
c2 = 2x2
c = √ ## 2x2
c = x √ # 2
In a 45°-45°-90° right
triangle the hypotenuse is √ " 2 times
the leg. If the hypotenuse is 6 units,
find the length of each leg.
The hypotenuse is √ # 2 times the leg, so
divide the length of the hypotenuse by √ # 2 .
a = 6 −
√ # 2
= 6 −
√ # 2 .
√ # 2 −
√ # 2
= 6 √ # 2
− 2
= 3 √ # 2 units
Exercises
Find x.
1. x
8
45°
45°
2.
x
45°3√2 3.
45°
4
x
4.
x x
18 5.
45°
16
x
x
6.
45°
24
x
√2
7. If a 45°-45°-90° triangle has a hypotenuse length of 12, find the leg length.
8. Determine the length of the leg of 45°-45°-90° triangle with a hypotenuse length of
25 inches.
9. Find the length of the hypotenuse of a 45°-45°-90° triangle with a leg length of
14 centimeters.
8-3
Example 1 Example 2
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Chapter 8 19 Glencoe Geometry
Study Guide and Intervention (continued)
Special Right Triangles
Properties of 30°-60°-90° Triangles The sides of a 30°-60°-90° right triangle also have a special relationship.
In a 30°-60°-90° right triangle the hypotenuse
is twice the shorter leg. Show that the longer leg is √ " 3 times
the shorter leg.
△ MNQ is a 30°-60°-90° right triangle, and the length of the hypotenuse
−−− MN is two times the length of the shorter side
−−− NQ .
Use the Pythagorean Theorem.
a2 = (2x)2 - x2 a2 = c2 - b2
a2 = 4x2 - x2 Multiply.
a2 = 3x2 Subtract.
a = √ $$ 3x2 Take the positive square root of each side.
a = x √ $ 3 Simplify.
In a 30°-60°-90° right triangle, the hypotenuse is 5 centimeters.
Find the lengths of the other two sides of the triangle.
If the hypotenuse of a 30°-60°-90° right triangle is 5 centimeters, then the length of the
shorter leg is one-half of 5, or 2.5 centimeters. The length of the longer leg is √ $ 3 times the
length of the shorter leg, or (2.5)( √ $ 3 ) centimeters.
Exercises
Find x and y.
1. x
y30°
60°1
2
2.
x
y
60°
8
3.
x
y
11
30°
4.
xy
30°
9√3
5.
xy60°
12
6. 60°
x
20y
7. An equilateral triangle has an altitude length of 36 feet. Determine the length of a side of the triangle.
8. Find the length of the side of an equilateral triangle that has an altitude length of 45 centimeters.
8-3
30°
60°
x2x
aExample 1
Example 2
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Chapter 8 20 Glencoe Geometry
8-3 Skills Practice
Special Right Triangles
Find x.
1.
45°
25
x
2.
45°
17x
3. 45°
48
x
4.
45°
100
x
5.
45°
100
x
6.
45°
88 x
7. Determine the length of the leg of 45°-45°-90° triangle with a hypotenuse length of 26.
8. Find the length of the hypotenuse of a 45°-45°-90° triangle with a leg length of
50 centimeters.
Find x and y.
9.
30°
x11
y
10.
60°x
8
y
√3 11. 30°
x
5
y
√3
12.
60°
x
30
y
13.
30°
x
21y
√3
14. 60°
x
52y
√3
15. An equilateral triangle has an altitude length of 27 feet. Determine the length of a side
of the triangle.
16. Find the length of the side of an equilateral triangle that has an altitude length of
11 √ " 3 feet.
Lesso
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Chapter 8 21 Glencoe Geometry
Find x.
1.
45°
14
x
2.
45°
45x
3. 45°
22
x
4.
45°
210
x
5.
45°
88
x
6.
45°
x5√2
Find x and y.
7.
30°
x9
y
8.
60°x
4
y
√3
9.
30°
x
20
y
10.
60°
x
98
y
11. Determine the length of the leg of 45°-45°-90° triangle with a hypotenuse length of 38.
12. Find the length of the hypotenuse of a 45°-45°-90° triangle with a leg length of
77 centimeters.
13. An equilateral triangle has an altitude length of 33 feet. Determine the length of a side
of the triangle.
14. BOTANICAL GARDENS One of the displays at a botanical garden
is an herb garden planted in the shape of a square. The square
measures 6 yards on each side. Visitors can view the herbs from a
diagonal pathway through the garden. How long is the pathway?6 yd 6 yd
6 yd
6 yd
Practice
Special Right Triangles
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Chapter 8 22 Glencoe Geometry
Word Problem Practice
Special Right Triangles
1. ORIGAMI A square piece of paper
150 millimeters on a side is folded in
half along a diagonal. The result is a
45°-45°-90° triangle. What is the length
of the hypotenuse of this triangle?
2. ESCALATORS A 40-foot-long escalator
rises from the first floor to the second
floor of a shopping mall. The escalator
makes a 30° angle with the horizontal.
40 ft
30°
? ft
How high above the first floor is the
second floor?
3. HEXAGONS A box of chocolates shaped
like a regular hexagon is placed snugly
inside of a rectangular box as shown in
the figure.
If the side length of the hexagon is
3 inches, what are the dimensions of
the rectangular box?
4. WINDOWS A large stained glass
window is constructed from six 30°-60°-
90° triangles as shown in the figure.
3 m
What is the height of the window?
5. MOVIES Kim and Yolanda are
watching a movie in a movie theater.
Yolanda is sitting x feet from the screen
and Kim is 15 feet behind Yolanda.
30° 45°
15 ft x ft
The angle that Kim’s line of sight
to the top of the screen makes with
the horizontal is 30°. The angle that
Yolanda’s line of sight to the top of the
screen makes with the horizontal is 45°.
a. How high is the top of the screen in
terms of x?
b. What is x + 15
− x ?
c. How far is Yolanda from the screen?
Round your answer to the nearest
tenth.
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Chapter 8 23 Glencoe Geometry
Enrichment
√
1
1
1
3√
2
Constructing Values of Square RootsThe diagram at the right shows a right isosceles triangle with
two legs of length 1 inch. By the Pythagorean Theorem, the length
of the hypotenuse is √ # 2 inches. By constructing an adjacent right
triangle with legs of √ # 2 inches and 1 inch, you can create a segment
of length √ # 3 .
By continuing this process as shown below, you can construct a
“wheel” of square roots. This wheel is called the “Wheel of Theodorus”
after a Greek philosopher who lived about 400 B.C.
Continue constructing the wheel until you make a segment of
length √ ## 18 .
1
1
11
1
1
√2
√3
√5
√6
√4 = 2
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Chapter 8 24 Glencoe Geometry
Trigonometric Ratios The ratio of the lengths of two sides of a right triangle is called a trigonometric ratio. The three most common ratios are sine, cosine, and tangent, which are abbreviated sin, cos, and tan, respectively.
sin R = leg opposite ∠R
− hypotenuse
cos R = leg adjacent to ∠R
−− hypotenuse
tan R = leg opposite ∠R
−− leg adjacent to ∠R
= r − t =
s −
t =
r −
s
Find sin A, cos A, and tan A. Express each ratio as
a fraction and a decimal to the nearest hundredth.
sin A = opposite leg
− hypotenuse
cos A = adjacent leg
− hypotenuse
tan A = opposite leg
− adjacent leg
= BC
− BA
= AC
− AB
= BC
− AC
= 5 −
13 = 12
− 13
= 5 −
12
≈ 0.38 ≈ 0.92 ≈ 0.42
Exercises
Find sin J, cos J, tan J, sin L, cos L, and tan L. Express each ratio as a fraction
and as a decimal to the nearest hundredth if necessary.
1. 2. 3.
12
135
C
B
A
Study Guide and Intervention
Trigonometry
8-4
s
tr
T
S
R
Example
20
12
16
40
24
32
36
12√324√3
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Less
on
8-4
Chapter 8 25 Glencoe Geometry
Study Guide and Intervention (continued)
Trigonometry
Use Inverse Trigonometric Ratios You can use a calculator and the sine, cosine, or tangent to find the measure of the angle, called the inverse of the trigonometric ratio.
Use a calculator to find the measure of ∠T to the nearest tenth.
The measures given are those of the leg opposite ∠T and the
hypotenuse, so write an equation using the sine ratio.
sin T = opp
− hyp
sin T = 29 −
34
If sin T = 29
− 34
, then sin-1 29
− 34
= m∠T.
Use a calculator. So, m∠T ≈ 58.5.
Exercises
Use a calculator to find the measure of ∠T to the nearest tenth.
1.
34
14√3
2. 7
18
3.
87
34
4.
101
32
5.
67
10√3
6.
39
14√2
8-4
Example
34
29
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Chapter 8 26 Glencoe Geometry
Find sin R, cos R, tan R, sin S, cos S, and tan S.
Express each ratio as a fraction and as a decimal to the
nearest hundredth.
1. r = 16, s = 30, t = 34 2. r = 10, s = 24, t = 26
Use a special right triangle to express each trigonometric ratio as a fraction and
as a decimal to the nearest hundredth if necessary.
3. sin 30° 4. tan 45° 5. cos 60°
6. sin 60° 7. tan 30° 8. cos 45°
Find x. Round to the nearest hundredth if necessary.
9.
7 x
36°
10. 11.
Use a calculator to find the measure of ∠B to the nearest tenth.
12. 13. 14.
sR
S
T
rt
Skills Practice
Trigonometry
8-4
12x
15°
6
x°
18
15
x°
12
22
x°
19
12 x
63°
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Chapter 8 27 Glencoe Geometry
Find sin L, cos L, tan L, sin M, cos M, and tan M.
Express each ratio as a fraction and as a decimal to the
nearest hundredth.
1. ℓ = 15, m = 36, n= 39 2. ℓ = 12, m = 12 √ # 3 , n = 24
Find x. Round to the nearest hundredth.
3.
11
64°
x
4. 5.
41°
x
32
Use a calculator to find the measure of ∠B to the nearest tenth.
6.
8
14
7.
30
25
8.
9. GEOGRAPHY Diego used a theodolite to map a region of land for his
class in geomorphology. To determine the elevation of a vertical rock
formation, he measured the distance from the base of the formation to
his position and the angle between the ground and the line of sight to the
top of the formation. The distance was 43 meters and the angle was
36°. What is the height of the formation to the nearest meter?
29
29°x
36°
43m
ML
N
m
n
Practice
Trigonometry
8-4
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Chapter 8 28 Glencoe Geometry
Word Problem Practice
Trigonometry
1. RADIO TOWERS Kay is standing near
a 200-foot-high radio tower.
200 ft ? ft
49°
Use the information in the figure to
determine how far Kay is from the top of
the tower. Express your answer as a
trigonometric function.
2. RAMPS A 60-foot ramp rises from the
first floor to the second floor of a parking
garage. The ramp makes a 15° angle
with the ground.
? ft
60 ft
15°
How high above the first floor is the
second floor? Express your answer as a
trigonometric function.
3. TRIGONOMETRY Melinda and Walter
were both solving the same trigonometry
problem. However, after they finished
their computations, Melinda said the
answer was 52 sin 27° and Walter said
the answer was 52 cos 63°. Could they
both be correct? Explain.
4. LINES Jasmine draws line m on a
coordinate plane.
m
x
y
O-5
5
5
What angle does m make with the
x-axis? Round your answer to the
nearest degree.
5. NEIGHBORS Amy, Barry, and Chris live
on the same block. Chris lives up the
street and around the corner from Amy,
and Barry lives at the corner between
Amy and Chris. The three homes are the
vertices of a right triangle.
64°
Barry
Chris
Amy
26°
a. Give two trigonometric expressions
for the ratio of Barry’s distance from
Amy to Chris’ distance from Amy.
b. Give two trigonometric expressions
for the ratio of Barry’s distance from
Chris to Amy’s distance from Chris.
c. Give a trigonometric expression for
the ratio of Amy’s distance from
Barry to Chris’ distance from Barry.
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Lesso
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-4
Chapter 8 29 Glencoe Geometry
Enrichment
Sine and Cosine of AnglesThe following diagram can be used to obtain approximate values for the sine and cosine of angles from 0° to 90°. The radius of the circle is 1. So, the sine and cosine values can be read directly from the vertical and horizontal axes.
Find approximate values for sin 40° and
cos 40°. Consider the triangle formed by the segment
marked 40°, as illustrated by the shaded triangle at right.
sin 40° = a −
c ≈
0.64 −
1 or 0.64 cos 40° =
b −
c ≈
0.77 −
1 or 0.77
1. Use the diagram above to complete the chart of values.
x° 0° 10° 20° 30° 40° 50° 60° 70° 80° 90°
sin x° 0.64
cos x° 0.77
2. Compare the sine and cosine of two complementary angles (angles with a sum of 90°). What do you notice?
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
90°
0°
10°
20°
30°
40°
50°
60°
70°
80°
1
0
40°0.64
c = 1 unit
x°
b = cos x° 0.77 1
a = sin x°
Example
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Chapter 8 30 Glencoe Geometry
Study Guide and Intervention
Angles of Elevation and Depression
Angles of Elevation and Depression Many real-world problems that involve looking up to an object can be described in terms of an angle of elevation, which is the angle between an observer’s line of sight and a horizontal line.
When an observer is looking down, the angle of depression is the angle between the observer’s line of sight and a horizontal line.
The angle of elevation from point A to the top
of a cliff is 34°. If point A is 1000 feet from the base of the cliff,
how high is the cliff?
Let x = the height of the cliff.
tan 34° = x −
1000 tan =
opposite −
adjacent
1000(tan 34°) = x Multiply each side by 1000.
674.5 ≈ x Use a calculator.
The height of the cliff is about 674.5 feet.
Exercises
1. HILL TOP The angle of elevation from point A to the top of a hill is 49°. If point A is 400 feet from the base of the hill, how high is the hill?
2. SUN Find the angle of elevation of the Sun when a 12.5-meter-tall telephone pole casts an 18-meter-long shadow.
3. SKIING A ski run is 1000 yards long with a vertical drop of 208 yards. Find the angle of depression from the top of the ski run to the bottom.
4. AIR TRAFFIC From the top of a 120-foot-high tower, an air traffic controller observes an airplane on the runway at an angle of depression of 19°. How far from the base of the tower is the airplane?
18 m
12.5 m
Sun
?
400 ft
?
49°A
?
1000 ft
34°
x
angle ofelevation
line o
f sight
8-5
Example
Yline of s
ight
horizontal
angle ofdepression
120 ft
?
19°
208 yd
?
1000 yd
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Chapter 8 31 Glencoe Geometry
Two Angles of Elevation or Depression Angles of elevation or depression to two different objects can be used to estimate distance between those objects. The angles from two different positions of observation to the same object can be used to estimate the height of the object.
To estimate the height of a garage,
Jason sights the top of the garage at a 42° angle
of elevation. He then steps back 20 feet and sites
the top at a 10° angle. If Jason is 6 feet tall,
how tall is the garage to the nearest foot?
△ ABC and △ ABD are right triangles. We can determine AB = x and CB = y, and DB = y + 20.
Use △ ABC. Use △ ABD.
tan 42° = x − y or y tan 42° = x tan 10° =
x −
y + 20 or (y + 20) tan 10° = x
Substitute the value for x from △ ABD in the equation for △ ABC and solve for y.
y tan 42° = (y + 20) tan 10°
y tan 42° = y tan 10° + 20 tan 10°
y tan 42° − y tan 10° = 20 tan 10°
y (tan 42° − tan 10°) = 20 tan 10°
y = 20 tan 10°
−− tan 42° - tan 10°
≈ 4.87
If y ≈ 4.87, then x = 4.87 tan 42° or about 4.4 feet. Add Jason’s height, so the garage is about 4.4 + 6 or 10.4 feet tall.
Exercises
1. CLIFF Sarah stands on the ground and sights the top of a steep cliff at a 60° angle of elevation. She then steps back 50 meters and sights the top of the steep cliff at a 30° angle. If Sarah is 1.8 meters tall, how tall is the steep cliff to the nearest meter?
2. BALLOON The angle of depression from a hot air balloon in the air to a person on the ground is 36°. If the person steps back 10 feet, the new angle of depression is 25°. If the person is 6 feet tall, how far off the ground is the hot air balloon?
Study Guide and Intervention (continued)
Angles of Elevation and Depression
8-5
Example
42°10°
x
y6 ft 20 ft
Building
60°30°
x
y1.8m 50m
Steepcliff
36°25°
x
y6 ft 10 ft
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Chapter 8 32 Glencoe Geometry
Name the angle of depression or angle of elevation in each figure.
1. 2.
3. 4.
5. MOUNTAIN BIKING On a mountain bike trip along the Gemini Bridges Trail in Moab,
Utah, Nabuko stopped on the canyon floor to get a good view of the twin sandstone
bridges. Nabuko is standing about 60 meters from the base of the canyon cliff, and the
natural arch bridges are about 100 meters up the canyon wall. If her line of sight is
5 metres above the ground, what is the angle of elevation to the top of the bridges?
Round to the nearest tenth degree.
6. SHADOWS Suppose the sun casts a shadow off a 35-foot building.
If the angle of elevation to the sun is 60°, how long is the shadow
to the nearest tenth of a foot?
7. BALLOONING Angie sees a hot air balloon in the
sky from her spot on the ground. The angle of
elevation from Angie to the balloon is 40°. If she steps
back 200 feet, the new angle of elevation is 10°.
If Angie is 5.5 feet tall, how far off the ground
is the hot air balloon?
8. INDIRECT MEASUREMENT Kyle is at the end
of a pier 30 feet above the ocean. His eye level is
3 feet above the pier. He is using binoculars to
watch a whale surface. If the angle of depression
of the whale is 20°, how far is the whale from
Kyle’s binoculars? Round to the nearest tenth foot.
whale water level
20°
Kyle’s eyes
pier3 ft
30 ft
60°?
35 ft
Z
P
W
R
D
A
C
B
T
W
R
S
F
T
L
S
Skills Practice
Angles of Elevation and Depression
8-5
40°
10°x
y5.5 ft 200 ft
Balloon
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Chapter 8 33 Glencoe Geometry
Name the angle of depression or angle of elevation in each figure.
1. 2.
3. WATER TOWERS A student can see a water tower from the closest point of the soccer
field at San Lobos High School. The edge of the soccer field is about 110 feet from the
water tower and the water tower stands at a height of 32.5 feet. What is the angle of
elevation if the eye level of the student viewing the tower from the edge of the soccer
field is 6 feet above the ground? Round to the nearest tenth.
4. CONSTRUCTION A roofer props a ladder against a wall so that the top of the ladder
reaches a 30-foot roof that needs repair. If the angle of elevation from the bottom of the
ladder to the roof is 55°, how far is the ladder from the base of the wall? Round your
answer to the nearest foot.
5. TOWN ORDINANCES The town of Belmont restricts the height
of flagpoles to 25 feet on any property. Lindsay wants to
determine whether her school is in compliance with the regulation.
Her eye level is 5.5 feet from the ground and she stands 36 feet
from the flagpole. If the angle of elevation is about 25°,
what is the height of the flagpole to the nearest tenth?
6. GEOGRAPHY Stephan is standing on the ground by
a mesa in the Painted Desert. Stephan is 1.8 meters
tall and sights the top of the mesa at 29°. Stephan steps
back 100 meters and sights the top at 25°. How tall is
the mesa?
7. INDIRECT MEASUREMENT Mr. Dominguez is standing
on a 40-foot ocean bluff near his home. He can see his two
dogs on the beach below. If his line of sight is 6 feet above
the ground and the angles of depression to his dogs are
34° and 48°, how far apart are the dogs to the nearest foot?
48°34°
40 ft
6 ft
Mr. Dominguez
bluff
25°
5.5 ft
36 ft
x
R
M
P
L
T
Y
R
Z Y
Practice
Angles of Elevation and Depression
8-5
29°25°
x
y1.8 m
100 m
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Chapter 8 34 Glencoe Geometry
Word Problem Practice
Angles of Elevation and Depression
1. LIGHTHOUSES Sailors on a ship at sea
spot the light from a lighthouse. The
angle of elevation to the light is 25°.
25°
30m
?
The light of the lighthouse is 30 meters
above sea level. How far from the shore
is the ship? Round your answer to the
nearest meter.
2. RESCUE A hiker dropped his backpack
over one side of a canyon onto a ledge
below. Because of the shape of the cliff,
he could not see exactly where it landed.
32°
115 ft
? ft
Backpack
From the other side, the park ranger
reports that the angle of depression to
the backpack is 32°. If the width of the
canyon is 115 feet, how far down did
the backpack fall? Round your answer
to the nearest foot.
3. AIRPLANES The angle of elevation to
an airplane viewed from the control
tower at an airport is 7°. The tower is
200 feet high and the pilot reports that
the altitude is 5200 feet. How far away
from the control tower is the airplane?
Round your answer to the nearest foot.
4. PEAK TRAM The Peak Tram in Hong
Kong connects two terminals, one at the
base of a mountain, and the other at the
summit. The angle of elevation of the
upper terminal from the lower terminal
is about 15.5°. The distance between the
two terminals is about 1365 meters.
About how much higher above sea level
is the upper terminal compared to the
lower terminal? Round your answer to
the nearest meter.
5. HELICOPTERS Jermaine and John are
watching a helicopter hover above the
ground.
48° 55°
(Not drawn to scale)
Helicopter
Jermaine John10 m
h
x
Jermaine and John are standing
10 meters apart.
a. Find two different expressions that
can be used to find the h, height of
the helicopter.
b. Equate the two expressions you found
for Exercise a to solve for x. Round
your answer to the nearest
hundredth.
c. How high above the ground is the
helicopter? Round your answer to the
nearest hundredth.
8-5
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Chapter 8 35 Glencoe Geometry
Enrichment
Best Seat in the House
Most people want to sit in the best seat in the movie theater. The best seat could be
defined as the seat that allows you to see the maximum amount of screen. The picture
below represents this situation.
To determine the best seat in the house, you want to find what value of x allows you to see
the maximum amount of screen. The value of x is how far from the screen you should sit.
1. To maximize the amount of screen viewed, which angle value needs to be maximized?
Why?
2. What is the value of a if x = 10 feet?
3. What is the value of a if x = 20 feet?
4. What is the value of a if x = 25 feet?
5. What is the value of a if x = 35 feet?
6. What is the value of a if x = 55 feet?
7. Which value of x gives the greatest value of a? So, where is the best seat in the movie
theater?
a˚
b˚
c˚
x ft
screen
40 ft
10 ft
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Chapter 8 36 Glencoe Geometry
Study Guide and Intervention
The Law of Sines and Law of Cosines
The Law of Sines In any triangle, there is a special relationship between the angles of the triangle and the lengths of the sides opposite the angles.
Find b. Round to the
nearest tenth.
45°
3074°
b
B
AC
sin C
− c =
sin B −
b Law of Sines
sin 45°
− 30
= sin 74°
− b m∠C = 45, c = 30, m∠B = 74
b sin 45° = 30 sin 74° Cross Products Property
b = 30 sin 74°
− sin 45°
Divide each side by sin 45°.
b ≈ 40.8 Use a calculator.
Find d. Round to the
nearest tenth.
By the Triangle Angle-Sum Theorem, m∠E = 180 - (82 + 40) or 58.
82° 40°
24
d
sin D
− d =
sin E −
e Law of Sines
sin 82°
− d =
sin 58° −
24 m∠D = 82, m∠E = 58,
e = 24
24 sin 82° = d sin 58° Cross Products Property
24 sin 82° −
sin 58° = d Divide each side by sin 58°.
d ≈ 28.0 Use a calculator.
Exercises
Find x. Round to the nearest tenth.
1.
80°
40°
x
12
2.
52°
91°
x
20 3.
16x
84°
34°
4.
17° 72°
x25
5.
89°
80°x
12 6.
60°
35°
35 x
8-6
Example 1 Example 2
Law of Sines sin A
− a =
sin B −
b =
sin C −
c
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Chapter 8 37 Glencoe Geometry
Study Guide and Intervention (continued)
The Law of Sines and Law of Cosines
The Law of Cosines Another relationship between the sides and angles of any triangle is called the Law of Cosines. You can use the Law of Cosines if you know three sides of a triangle or if you know two sides and the included angle of a triangle.
Law of Cosines
Let △ABC be any triangle with a, b, and c representing the measures of the sides opposite
the angles with measures A, B, and C, respectively. Then the following equations are true.
a2 = b2 + c2 - 2bc cos A b2 = a2 - c2 - 2ac cos B c2 = a2 + b2 - 2ab cos C
Find c. Round to the nearest tenth.
c2 = a2 + b2 - 2ab cos C Law of Cosines
c2 = 122 + 102 - 2(12)(10)cos 48° a = 12, b = 10, m∠C = 48
c = √ %%%%%%%%%%%% 122 + 102 - 2(12)(10)cos 48° Take the square root of each side.
c ≈ 9.1 Use a calculator.
Find m∠A. Round to the nearest degree.
a2 = b2 + c2 - 2bc cos A Law of Cosines
72 = 52 + 82 - 2(5)(8) cos A a = 7, b = 5, c = 8
49 = 25 + 64 - 80 cos A Multiply.
-40 = -80 cos A Subtract 89 from each side.
1 − 2 = cos A Divide each side by -80.
cos-1 1 − 2 = A Use the inverse cosine.
60° = A Use a calculator.
Exercises
Find x. Round angle measures to the nearest degree and side measures to the
nearest tenth.
1. 2. 3.
4. 5. 6.
8-6
Example 1
Example 2
48°10 12
c
C
BA
12
14 x
62°
12
11
10
x°
16
24
18
x°
25
20x
82°
18
28
x
59°
18
15
15
x°
8
5
7
C
B
A
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Chapter 8 38 Glencoe Geometry
Find x. Round angle measures to the nearest degree and side lengths to the
nearest tenth.
1.
28
35° 48°
x
2.
18
46°
17°
x
3.
38
x
86°
51°
4. 5. 6.
7. 8. 9.
10. 11. 12.
13. 14. 15.
16. Solve the triangle. Round angle measures to the nearest degree.
Skills Practice
The Law of Sines and Law of Cosines
8-6
x8
73°
80°
x
34
41°
88°
x
1283°
60°
18
x
104°
22°
x 27
54°
93°
12 16
11
x°
17
18
23
29
111°
34
x
16
89°
12
x 13
103°
20
x
13
35
x
96° 28
x°26
30
25
28°
x20
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Chapter 8 39 Glencoe Geometry
Practice
The Law of Sines and Law of Cosines
13. INDIRECT MEASUREMENT To find the distance from the edge
of the lake to the tree on the island in the lake, Hannah set up
a triangular configuration as shown in the diagram. The distance
from location A to location B is 85 meters. The measures of the
angles at A and B are 51° and 83°, respectively. What is the
distance from the edge of the lake at B to the tree on the island at C?
A
C
B
Find x. Round angle measures to the nearest degree and side lengths to the
nearest tenth.
1. 2. 3.
4. 5. 6.
7. 8. 9.
10. 11. 12.
14
x
67°
14°
127
x
42°
61°
x
5.8
83°
73°
19.1
x
56°
34°
9.6
x
43°
123° 4.3
x
85°
64°
40
12
37°
x 23
28
37°
x
15
17
62°
x
28.414.5
21.7
x°
89°
14
15
x
1410
9.6
x°
8-6
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Chapter 8 40 Glencoe Geometry
Word Problem Practice
The Law of Sines and Law of Cosines
1. ALTITUDES In triangle ABC, the
altitude to side −−
AB is drawn.
A
ab
B
C
Give two expressions for the length of
the altitude in terms of a, b, and the
sine of the angles A and B.
2. MAPS Three cities form the vertices of
a triangle. The angles of the triangle are
40°, 60°, and 80°. The two most distant
cities are 40 miles apart. How close are
the two closest cities? Round your
answer to the nearest tenth of a mile.
3. STATUES Gail was visiting an art
gallery. In one room, she stood so that
she had a view of two statues, one of a
man, and the other of a woman. She was
40 feet from the statue of the woman,
and 35 feet from the statue of the man.
The angle created by the lines of sight to
the two statues was 21°. What is the
distance between the two statues?
Round your answer to the nearest tenth.
4. CARS Two cars start moving from the
same location. They head straight, but
in different directions. The angle
between where they are heading is 43°.
The first car travels 20 miles and the
second car travels 37 miles. How far
apart are the two cars? Round your
answer to the nearest tenth.
5. ISLANDS Oahu
is a Hawaiian
Island. Off of the
coast of Oahu,
there is a very
tiny island
known as
Chinaman’s Hat.
Keoki and Malia
are observing
Chinaman’s Hat from locations
5 kilometers apart. Use the information
in the figure to answer the following
questions.
a. How far is Keoki from Chinaman’s
Hat? Round your answer to the
nearest tenth of a kilometer.
b. How far is Malia from Chinaman’s
Hat? Round your answer to the
nearest tenth of a kilometer.
Ka’a’awa
Keoki Chinaman’s
Hat
5 km
Malia
Kahalu’u
22.3°
49.5°
8-6
Statue of
a man
Statue of a woman
35 ft
40 ft
21˚
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Chapter 8 41 Glencoe Geometry
Enrichment
Identities
An identity is an equation that is true for all values of the variable for which both sides are defined. One way to verify an identity is to use a right triangle and the definitions for trigonometric functions.
Verify that (sin A)2 + (cos A)2 = 1 is an identity.
(sin A)2 + (cos A)2 = ( a − c ) 2 + (
b − c ) 2
= a2+ b2
− c2
= c2
− c2
= 1
To check whether an equation may be an identity, you can test several values. However, since you cannot test all values, you cannot be certain that the equation is an identity.
Test sin 2x = 2 sin x cos x to see if it could be an identity.
Try x = 20. Use a calculator to evaluate each expression.
sin 2x = sin 40 2 sin x cos x = 2 (sin 20)(cos 20) ≈ 0.643 ≈ 2(0.342)(0.940) ≈ 0.643
Since the left and right sides seem equal, the equation may be an identity.
Exercises
Use triangle ABC shown above. Verify that each equation is an identity.
1. cos A
− sin A
= 1 −
tan A 2.
tan B −
sin B = 1
− cos B
3. tan B cos B = sin B 4. 1 - (cos B)2 = (sin B)2
Try several values for x to test whether each equation could be an identity.
5. cos 2x = (cos x)2 - (sin x)2 6. cos (90 - x) = sin x
B
A C
ca
b
8-6
Example 1
Example 2
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Chapter 8 42 Glencoe Geometry
Graphing Calculator Activity
Solving Triangles Using the Law of Sines or Cosines
You can use a calculator to solve triangles using the Law of Sines or Cosines.
Solve △ABC if a = 6, b = 2, and c = 7.5.
Use the Law of Cosines.
a2 = b2 + c2 - 2bc cos A
62 = 22 + 7.52 - 2(2)(7.5) cos A
m∠A = cos-1 62 - 22 - 7.52
− -2(2)(7.5)
Use your calculator to find the measure of ∠A.
Keystrokes: 2nd [COS-1] ( 6 x2 — 2 x2 — 7.5 x2 ) ÷
Enter: ( (–) 2 3 2 3 7.5 ) ) ENTER 36.06658826
So m∠A ≈ 36. Use the Law of Sines and your calculator to find m∠B.
sin A
− a =
sin B −
b
sin 36
− 6 ≈
sin B −
2
m∠B ≈ sin-1 2 sin 36°
− 6
Keystrokes: 2nd [SIN-1] ( 2 SIN 36 ) ÷ 6 ) ENTER 11.29896425
So m∠B ≈ 11. By the Triangle Angle-Sum Theorem, m∠C ≈ 180 - (36 + 11) or 133.
Exercises
Solve each △ABC. Round measures of sides to the nearest tenth
and measures of angles to the nearest degree.
1. a = 9, b = 14, c = 12
2. m∠C = 80, c = 9, m∠A = 40
3. m∠B = 45, m∠C = 56, a = 2
4. a = 5.7, b = 6, c = 5
5. a = 11, b = 15, c = 21
8-6
Example
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Chapter 8 43 Glencoe Geometry
Geometric Vector Operations A vector is a directed segment representing a quantity that has both magnitude, or length, and direction. For example, the speed and direction of an
airplane can be represented by a vector. In symbols, a vector is written as ! AB , where A is the
initial point and B is the endpoint, or as ! v . The sum of two vectors is called the resultant.
Subtracting a vector is equivalent to adding its opposite. The resultant of two vectors can be found using the parallelogram method or the triangle method.
Copy the vectors to find a -
b . a#
b#
Method 1: Use the parallelogram method.
Copy ! a and -
! b with the same initial
point.
a
-b
Complete the parallelogram.
a
-b
Draw the diagonal of the
parallelogram from the initial point.
a
a
-b
-b
Method 2: Use the triangle method.
Copy ! a .
a
Place the initial point of - ! b at the
terminal point of ! a .
a
-b
Draw the vector from the initial
point of ! a to the terminal point
of - ! b .
a
-b
a - b
Exercises
Copy the vectors. Then find each sum or difference.
1. ! c +
! d 2. w -
! z
3. ! a -
! b 4.
! r +
! t
##
Example
c# d#
w#
z#
r#
t#
a#
b#
Study Guide and Intervention
Vectors
8-7
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Chapter 8 44 Glencoe Geometry
Study Guide and Intervention (continued)
Vectors
8-7
Vectors on the Coordinate Plane A vector in standard position has its initial point at (0, 0) and can be represented by the ordered
pair for point B. The vector at the right can be expressed as v = ⟨5, 3⟩.
You can use the Distance Formula to find the magnitude | AB | of a vector. You can describe the direction of a vector
by measuring the angle that the vector forms with the positive x-axis or with any other horizontal line.
Find the magnitude and direction of a = ⟨3, 5⟩.
Find the magnitude.
a = √ &&&&&&&&& (x
2 - x
1)2 + (y
2 - y
1)2
= √ &&&&&&&& (3 - 0)2 + (5 - 0)2
= √ && 34 or about 5.8
To find the direction, use the tangent ratio.
tan θ = 5 −
3 The tangent ratio is opposite over adjacent.
m∠θ ≈ 59.0 Use a calculator.
The magnitude of the vector is about 5.8 units and its direction is 59°.
Exercises
Find the magnitude and direction of each vector.
1. b = ⟨-5, 2⟩ 2. c = ⟨-2, 1⟩
3. d = ⟨3, 4⟩ 4. m = ⟨5, -1⟩
5. r = ⟨-3, -4⟩ 6. v = ⟨-4, 1⟩
x
y
(3, 5)
x
y
(5, 3)
(0, 0)
Example
Distance Formula
(x1, y
1) = (0, 0) and (x
2, y
2) = (3, 5)
Simplify.
Lesso
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-7
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Chapter 8 45 Glencoe Geometry
Skills Practice
Vectors
8-7
Use a ruler and a protractor to draw each vector. Include a scale on each
diagram.
1. a = 20 meters per second 2. b = 10 pound of force at
60° west of south 135° to the horizontal
Copy the vectors to find each sum or difference.
3. a + z 4. t -
r
Write the component form of each vector.
5. 6.
Find the magnitude and direction of each vector.
7. m = ⟨2, 12⟩ 8. p = ⟨3, 10⟩
9. k = ⟨-8, -3⟩ 10. f = ⟨-5, 11⟩
Find each of the following for a = ⟨2, 4⟩,
b = ⟨3, -3⟩ , and
c = ⟨4, -1⟩. Check your
answers graphically.
x
y
O
B(–2, 2)
D(3, –3)
x
y
O
C(1, –1)
E(4, 3)
a
z
r
t
N
E
S
W
1 cm : 5 m
60°a
1 in : 10 lb
135°b
Overmatter
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Chapter 8 46 Glencoe Geometry
Practice
Vectors
Use a ruler and a protractor to draw each vector. Include a scale on each
diagram.
1. v = 12 Newtons of force at 2.
w = 15 miles per hour
40° to the horizontal 70° east of north
Copy the vectors to find each sum or difference.
3. p +
r 4.
a -
b
5. Write the component form of AB .
Find the magnitude and direction of each vector.
6. t = ⟨6, 11⟩ 7.
g = ⟨9, -7⟩
Find each of the following for a = ⟨-1.5, 4⟩,
b = ⟨7, 3⟩, and
c = ⟨1, -2⟩. Check your
answers graphically.
8. 2 a +
b 9. 2
c -
b
10. AVIATION A jet begins a flight along a path due north at 300 miles per hour. A wind
is blowing due west at 30 miles per hour.
a. Find the resultant velocity of the plane.
b. Find the resultant direction of the plane.
8-7
x
y
O
K(–2, 4)
L(3, –4)
p%r%a% b%
1 cm : 4 N 40°
V
N
ES
W
1 in : 10 mi
70°
w
b 2a
x
y
−4
4
8
12
4−4 8 12
2a + b b
y
−4
−8
4
8
x
−8−12 4
2c - b
2c
Lesso
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-7
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Chapter 8 47 Glencoe Geometry
Word Problem Practice
Vectors
8-7
1. WIND The vector v represents the speed
and direction that the wind is blowing.
Suddenly the wind picks up and doubles
its speed, but the direction does not
change. Write an expression for a vector
that describes the new wind velocity in
terms of v.
2. SWIMMING Jan is swimming in a
triathlon event. When the ocean water
is still, her velocity can be represented
by the vector ⟨2, 1⟩ miles per hour.
During the competition, there was a
fierce current represented by the
vector ⟨–1, –1⟩ miles per hour. What
vector represents Jan’s velocity during
the race?
3. POLYGONS Draw a regular polygon
around the origin. For each side of the
polygon, associate a vector whose
magnitude is the length of the
corresponding side and whose direction
points in the clockwise motion around
the origin. What vector represents the
sum of all these vectors? Explain.
4. BASEBALL Rick is in the middle of a
baseball game. His teammate throws
him the ball, but throws it far in front of
him. He has to run as fast as he can to
catch it. As he runs, he knows that as
soon as he catches it, he has to throw it
as hard as he can to the teammate at
home plate. He has no time to stop. In
the figure, x is the vector that represents
the velocity of the ball after Rick throws
it and v represents Rick’s velocity
because he is running. Assume that Rick
can throw just as hard when running as
he can when standing still.
Homeplate
Rick
x
v$
$
$X
$V
a. What vector would represent the
velocity of the ball if Rick threw it the
same way but he was standing still?
b. The angle between x and
v is 89°. By
running, did it help Rick get the ball
to home plate faster than he would
have normally been able to if he were
standing still?
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Chapter 8 48 Glencoe Geometry
Enrichment8-7
Dot Product
The dot product of two vectors represents how much the vectors point in the direction of each other. If
v is a vector represented by ⟨a, b⟩ and
u is a vector represented by ⟨c, d⟩, the
formula to find the dot product is:
v ·
u = ac + bd
Look at the following example:
Graph the vectors and find the dot product of v and
u if
v = ⟨3, -1⟩ and
u = ⟨2, 5⟩.
v ·
u = (3)(2) + (-1)(5) or 1
Graph the vectors and find the dot products.
1. v = ⟨2, 1⟩ and
u = ⟨-4, 2⟩ 2.
v = ⟨3, -2⟩ and
u = ⟨1, 4⟩
The dot product is The dot product is
3. v = ⟨0, 3⟩ and
u = ⟨2, 4⟩ 4.
v = ⟨-1, 4⟩ and
u = ⟨-4, 2⟩
The dot product is The dot product is
5. Notice the angle formed by the two vectors and the corresponding dot product. Is there any relationship between the type of angle between the two vectors and the sign of the dot product? Make a conjecture.
y
xO
y
xO
y
xO
y
xO
y
xO
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Assessment
Chapter 8 49 Glencoe Geometry
8
7.
Read each question. Then fill in the correct answer.
Record your answer in the blank.
For gridded response questions, also enter your answer in the grid by writing
each number or symbol in a box. Then fill in the corresponding circle for that
number or symbol.
Record your answers for Question 15 on the back of this paper.
Multiple Choice
Extended Response
Short Response/Gridded Response
1. A B C D
2. F G H J
3. A B C D
4. F G H J
5. A B C D
6. F G H J
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
. . . . .
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
7. (grid in)
8.
9.
10. (grid in)
11.
12.
13.
14.
10.
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
. . . . .
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
Student Recording SheetUse this recording sheet with pages 610–611 of the Student Edition.
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Chapter 8 50 Glencoe Geometry
8 Rubric for Scoring Extended-Response
General Scoring Guidelines
• If a student gives only a correct numerical answer to a problem but does not show how
he or she arrived at the answer, the student will be awarded only 1 credit. All extended-
response questions require the student to show work.
• A fully correct answer for a multiple-part question requires correct responses for all
parts of the question. For example, if a question has three parts, the correct response to
one or two parts of the question that required work to be shown is not considered a fully
correct response.
• Students who use trial and error to solve a problem must show their method. Merely
showing that the answer checks or is correct is not considered a complete response for
full credit.
Exercise 15 Rubric
Score Specific Criteria
4The values of x = 10, y = 16, and z = 12.8 are found using appropriate proportions based upon students knowledge of right triangle geometric mean theorems.
3A generally correct solution, but may contain minor flaws in reasoning or computation.
2 A partially correct interpretation and/or solution to the problem.
1 A correct solution with no evidence or explanation.
0An incorrect solution indicating no mathematical understanding of the concept or task, or no solution is given.
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Assessment
SCORE
NAME DATE PERIOD
Chapter 8 Quiz 2(Lessons 8-3 and 8-4)
8 SCORE
Chapter 8 51 Glencoe Geometry
8 Chapter 8 Quiz 1(Lessons 8-1 and 8-2)
For Questions 1 and 2, find x.
1. 2.
For Questions 3 and 4, find x to the nearest tenth.
3. 4.
5. A rectangle has a diagonal 20 inches long that forms angles of
60° and 30° with the sides. Find the perimeter of the rectangle.
6. Find sin 52°. Round to the nearest ten-thousandth.
7. If cos A = 0.8945, find m∠A to the nearest degree.
8. The distance along a hill is 24 feet. If the land slopes uphill at
an angle of 8°, find the vertical distance from the top to the
bottom of the hill. Round to the nearest tenth.
1. Find the geometric mean between 12 and 16.
For Questions 2 and 3, find x and y.
2.
5 12
xy
3.
8
9
x
y
4. Find x. 4
11
x
5. The measures of the sides of a triangle are 19, 15, and 13. Use
the Pythagorean Theorem to classify the triangle as acute,
obtuse, or right.
1.
2.
3.
4.
5.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
9. Use a calculator to find the
measure of ∠T to the
nearest tenth.
10. Use a calculator to
find the measure
of ∠T to the nearest
tenth.
17
31°
x
13
11
x°
45°
6x
60° 30°
6x
24
9
7121√3
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SCORE
Chapter 8 52 Glencoe Geometry
Chapter 8 Quiz 3(Lessons 8-5 and 8-6)
Chapter 8 Quiz 4(Lesson 8-7)
1. Name the angle of elevation in the figure.
2. Find x to the nearest tenth.
3. Solve △ABC. Round angle measures to the nearest degree and side measures to the nearest tenth.
4. Solve △RST. Round your answers to the nearest degree.
5. A squirrel, 37 feet up in a tree, sees a dog 29 feet from the base of the tree. Find the measure of the angle of depression to the nearest degree.
22°76°
10x
P
S
Q
R
Find the magnitude and direction of each vector.
1. " RT : R(14, 6) and T(−5, −10)
2. " PQ : P(20, −16) and Q(−9, −4)
Copy the vectors to find each sum or difference.
3. " c + " d 4.
" k -
" m
5. Lidia is rollerblading south at a velocity of 9 miles per hour. The wind is blowing 3 miles per hour in the opposite direction. What is Lidia’s resultant velocity and direction?
1.
2.
3.
4.
5.
8
48 61
76T S
R
8
49°
11 18
A C
B
d
c
m%k%
1.
2.
3.
4.
5.
Assessment
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Chapter 8 53 Glencoe Geometry
Chapter 8 Mid-Chapter Test(Lessons 8-1 through 8-4)
Part II
For Questions 6–8, find x and y.
6. 7.
8.
9. The measures of the sides of a triangle are 56, 90, and 106.
Use the Pythagorean Theorem to classify the triangle as acute,
obtuse, or right.
10. Guy wires 80 feet long support a 65-foot tall telephone pole. To
the nearest degree, what angle will the wires make with the
ground?
6.
7.
8.
9.
10.
1.
2.
3.
4.
5.
1. Find the geometric mean between 7 and 9.
A 3 √ " 7 B 16 C 8 D 2
2. Find x.
F 6 √ " 6 H 6 √ " 55
G √ " 2
− 5
J √ " 23
− 5
3. Find sin C.
A √ " 2 C √ " 23
− √ " 2
B √ " 2 −
5 D
√ " 23 −
5
4. Find x to the nearest tenth.
F 14 H 18.4
G 21.1 J 32.2
5. Find y to the nearest degree.
A 145 C 45
B 60 D 3527
19
y°
28
49°x
5√2
√"23B C
A
9
24
24
x
8
3
6 y
x
30°
60°
45°
8√3
y
x
60° 45°
20 y
x
Part I Write the letter for the correct answer in the blank at the right of each question.
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Chapter 8 54 Glencoe Geometry
Chapter 8 Vocabulary Test
Choose the correct word to complete each sentence.
1. The square root of the product of two numbers is the (geometric mean, or Pythagorean triple) of the numbers.
2. A group of three whole numbers that satisfy the equation a2 + b2 = c2, where c is the greatest number, is called a (trigonometric ratio, or Pythagorean triple).
Write whether each sentence is true or false. If false,
replace the underlined word or number to make a true
sentence.
3. The ratio of the lengths of any two sides of a right triangle is called a geometric mean.
4. An angle between the line of sight and the horizontal when anobserver looks upward is called a angle of elevation.
Choose from the terms above to complete each sentence.
5. An angle between the line of sight and the horizontal when an observer looks downward is called a(n) ? .
6. The word ? is derived from the Greek term for triangle and measure.
7. In a right triangle, the ? of an angle can be found by dividing the length of the opposite leg by the length of the triangle’s hypotenuse.
8. In a right triangle, the ? of an angle can be found by dividing the length of the opposite leg by the length of the adjacent leg.
Define each term in your own words.
9. component form
10. Pythagorean Theorem
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
angle of depression
angle of elevation
component form
cosine
direction
geometric mean
inverse cosine
inverse sine
inverse tangent
Law of Cosines
Law of Sines
magnitude
Pythagorean Triple
resultant
sine
standard position
tangent
trigonometric ratio
trigonometry
vector
8
Assessment
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Chapter 8 55 Glencoe Geometry
Chapter 8 Test, Form 1
Write the letter for the correct answer in the blank at the right of each question.
1. Find the geometric mean between 20 and 5.
A 100 B 50 C 12.5 D 10
2. Find x in △ABC.
F 8 H √ # 20
G 10 J 64
3. Find x in △PQR.
A 13 C 16
B 15 D √ # 60
4. Find x in △STU.
F 2 H √ # 32
G 8 J √ ## 514
5. Which set of measures could represent the lengths of the sides of a right
triangle?
A 2, 3, 4 C 8, 10, 12
B 7, 11, 14 D 9, 12, 15
6. Find x in △DEF.
F 6 H 6 √ # 3
G 6 √ # 2 J 12
7. Find y in △XYZ.
A 7.5 √ # 3 C 15
B 15 √ # 3 D 30
8. The length of the sides of a square is 10 meters. Find the length of the
diagonals of the square.
F 10 m H 10 √ # 3 m
G 10 √ # 2 m J 20 m
9. Find x in △HJK.
A 5 √ # 2 C 10
B 5 √ # 3 D 15
10. Find x in △ABC.
F 25 H 25 √ # 3
G 25 √ # 2 J 100
D E
F
6
6x
S T
U
15
17x
P Q
R
12
5x
A B
C 4
16x
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
8
A C
B
60° 30°
50
x
H J
K
60°
30°
5
x
X Y
Z
15√245°
45°
y
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Chapter 8 56 Glencoe Geometry
Chapter 8 Test, Form 1 (continued)
11. Find x to the nearest tenth.
A 7.3 C 18.4
B 17.3 D 47.1
12. Find the measure of the angle of elevation of the Sun when a pole 25 feet tall casts a shadow 42 feet long.
F 30.8° G 36.5° H 53.5° J 59.2°
13. Which is the angle of depression in the figure at the right?
A ∠AOT C ∠TOB
B ∠AOB D ∠BTO
14. Find y in △XYZ if m∠Y = 36, m∠X = 49, and x = 12. Round to the nearest hundreth.
F 0.04 G 9.35 H 14.80 J 15.41
15. To find the distance between two points, A and B, on opposite sides of a river, a surveyor measures the distance from A to C as 200 feet, m∠A = 72, and m∠B = 37. Find the distance from A to B. Round your answer to the nearest tenth.
A 77.4 ft B 201.2 ft C 250.4 ft D 314.2 ft
16. In △ABC, m∠A = 40, m∠C = 115, and b = 8. Find a to the nearest tenth.
F 12.2 G 11.3 H 5.7 J 5.3
17. Find the length of the third side of a triangular garden if two sides measure 8 feet and 12 feet and the included angle measures 50.
A 7.8 ft B 9.2 ft C 14.4 ft D 146.3 ft
18. In △DEF, d = 20, e = 25, and f = 30. Find m∠F to the nearest degree.
F 83° G 76° H 56° J 47°
19. Find the component form of # AB with A(2, 3) and B(−4, 6).
A ⟨−2, 9⟩ B ⟨2, −9⟩ C ⟨−6, 3⟩ D ⟨6, −3⟩
20. Find the magnitude of # AB with A(3, 4) and B(−1, 7).
F ⟨4, −3⟩ G 5 H √ ** 13 J 25
Bonus In △ABC, a = 50, b = 48, and c = 40. Find m∠ A to the nearest degree.
A B
C
AO
B T
67° 20
x
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
B:
8
Assessment
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Chapter 8 57 Glencoe Geometry
Chapter 8 Test, Form 2A
Write the letter for the correct answer in the blank at the right of each question.
1. Find the geometric mean between 7 and 12.
A 5 C √ " 19
B 9.5 D 2 √ " 21
2. In △PQR, RS = 4 and QS = 6. Find PS.
F 2 H √ " 10
G 5 J 2 √ " 6
3. Find x.
A 3 √ " 2 C 4.5
B √ " 14 D 3
4. Find y.
F 12 H 9
G 11 J 2
5. Find the length of the hypotenuse of a right triangle with legs that measure
5 and 7.
A 12 C √ " 35
B √ " 24 D √ " 74
6. Find x.
F 3 H 4 √ " 3
G 4 J 2 √ " 5
7. Which set of measures could represent the lengths of the sides of a right
triangle?
A 9, 40, 41 C 7, 8, 15
B 8, 30, 31 D √ " 2 , √ " 3 , √ " 6
8. Find c.
F 7 H 7 √ " 3
G 7 √ " 2 J 14
9. Find the perimeter of a square if the length of its diagonal is 12 inches.
Round to the nearest tenth.
A 8.5 in. C 48 in.
B 33.9 in. D 67.9 in.
10. Find x.
F 4 H 4 √ " 3
G 4 √ " 2 J 8 √ " 3 x
88
60°
45°
7c
x66
8
3
y 6
2 7
x
P Q
S
R4
6
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
8
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Chapter 8 58 Glencoe Geometry
Chapter 8 Test, Form 2A (continued)
11. Find x to the nearest tenth.
A 5.8 C 8.1
B 5.9 D 17.3
12. Find x to the nearest degree.
F 56 H 34
G 45 J 29
13. If a 20-foot ladder makes a 65° angle with the ground, how many feet up a wall will it reach? Round your answer to the nearest tenth.
A 8.5 ft B 10 ft C 18.1 ft D 42.9 ft
14. A ship’s sonar finds that the angle of depression to a wreck on the bottom of the ocean is 12.5°. If a point on the ocean floor is 60 meters directly below the ship, how many meters is it from that point on the ocean floor to the wreck? Round your answer to the nearest tenth.
F 277.2 m G 270.6 m H 61.5 m J 13.3 m
15. Find the angle of elevation of the sun if a building 100 feet tall casts a shadow 150 feet long. Round to the nearest degree.
A 60° B 48° C 42° D 34°
16. When the Sun’s angle of elevation is 73°, a tree tilted at an angle of 5° from the vertical casts a 20-foot shadow on the ground. Find the length of the tree to the nearest tenth of a foot.
F 6.3 ft H 51.1 ft
G 19.2 ft J 219.4 ft
17. In △CDE, m∠C = 52, m∠D = 17, and e = 28.6. Find c to the nearest tenth.
A 77.1 B 49.1 C 24.1 D 18.4
18. In △PQR, p = 56, r = 17, and m∠Q = 110. Find q to the nearest tenth.
F 4076.2 G 63.8 H 52.6 J 3.1
19. Find the component form of $ CD with C(5,-7) and D(-3, 9).
A ⟨-2, 2⟩ B ⟨2, 2⟩ C ⟨8, -16⟩ D ⟨-8, 16⟩
20. A pilot is flying due east at a speed of 300 miles per hour and wind is blowing due north at 50 miles per hour. What is the magnitude of the resultant velocity of the plane?
F 300 mph G 350 mph H about 304 mph J 2500 mph
Bonus From a window 20 feet above the ground, the angle of elevation to the top of another building is 35°. The distance between the buildings is 52 feet. Find the height of the building to the nearest tenth of a foot.
73°
5°
20-foot shadow
✹
95
x°
10
36°
x
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
B:
8
Assessment
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Chapter 8 59 Glencoe Geometry
Chapter 8 Test, Form 2B
Write the letter for the correct answer in the blank at the right of each question.
1. Find the geometric mean between 9 and 11.
A 3 √ " 11 C 10
B 2 √ " 5 D 2
2. In △PQR, RS = 5 and QS = 8. Find PS.
F 3 H √ " 13
G 6.5 J 2 √ " 10
3. Find x.
A 5.5 C √ " 24
B √ " 11 D √ " 33
4. Find y.
F 4 H 8
G 5 J 9
5. Find the length of the hypotenuse of a right triangle whose legs measure
6 and 5.
A 11 C √ " 30
B √ " 11 D √ " 61
6. Find x.
F √ " 39 H 5 √ " 3
G 6 J 5
7. Which set of measures could represent the lengths of the sides of a right
triangle?
A 3 −
4 , 1,
5 −
4 C 7, 17, 24
B √ " 3 , √ " 5 , √ " 15 D 8, 15, 16
8. Find c.
F 18 H 9 √ " 2
G 9 √ " 3 J 9
9. Find the perimeter of a square if the length of its diagonal is 16 millimeters.
Round to the nearest tenth.
A 11.3 mm C 90.5 mm
B 45.3 mm D 128.0 mm
10. Find x.
F 6 H 6 √ " 3
G 6 √ " 2 J 12 √ " 3
x1212
60°
45°
9c
x
8
8
10
y
4
6
x
83
P Q
SR 5
8
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
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Chapter 8 60 Glencoe Geometry
Chapter 8 Test, Form 2B (continued)
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
11. Find x.
A 8.0 C 10.4
B 8.9 D 10.8
12. Find x to the nearest degree.
F 57 H 33
G 55 J 29
13. If a 24-foot ladder makes a 58° angle with the ground, how many feet up a wall will it reach? Round your answer to the nearest tenth.
A 38.4 ft B 20.8 ft C 20.4 ft D 12.7 ft
14. A ship’s sonar finds that the angle of depression to a wreck on the bottom of the ocean is 13.2°. If a point on the ocean floor is 75 meters directly below the ship, how many meters is it from that point on the ocean floor to the wreck? Round to the nearest tenth.
F 328.4 m G 319.8 m H 77.0 m J 17.6 m
15. Find the angle of elevation of the sun if a building 125 feet tall casts a shadow 196 feet long. Round to the nearest degree.
A 64° B 50° C 40° D 33°
16. When the sun’s angle of elevation is 76°, a tree tilted at an angle of 4° from the vertical casts an 18-foot shadow on the ground. Find the length of the tree to the nearest tenth of a foot.
F 250.4 ft H 17.7 ft
G 56.5 ft J 4.6 ft
17. In △ABC, m∠A = 46, m∠B = 105, and c = 19.8. Find a to the nearest tenth.
A 29.4 B 28.5 C 15.7 D 14.7
18. In △LMN, ℓ = 42, m = 61, and m∠N = 108. Find n to the nearest tenth.
F 7068.4 G 84.1 H 79.2 J 24.7
19. Find the component from of $ EF with E(-11,-3) and F(7,-4)
A ⟨-18, 1⟩ B ⟨18, -1⟩ C ⟨-4, -7⟩ D ⟨4, 7⟩
20. An eagle is traveling along a path due east at a rate of 50 miles per hour and the wind is blowing due north at 15 miles per hour. What is the magnitude of the resultant velocity of the eagle?
F 35 mph G 50 mph H 65 mph J about 52.2 mph
Bonus From a window 24 feet above the ground, the angle of elevation to the top of another building is 38°. The distance between the buildings is 63 feet. Find the height of the building to the nearest tenth of a foot.
76°
4°
18-foot shadow
✹
x°
116
x
12
42°
B:
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Assessment
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Chapter 8 61 Glencoe Geometry
Chapter 8 Test, Form 2C
1. Find the geometric mean between 2 and 5.
For Questions 2–5, find x.
2. 3.
4. 5.
6. Find x.
7. In parallelogram ABCD, AD = 4
and m∠D = 60. Find AF.
8. Find x and y.
9. Find x to the nearest tenth.
10. An A-frame house is 40 feet high and
30 feet wide. Find the measure of the
angle that the roof makes with the floor.
Round to the nearest degree.
11. A 30-foot tree casts a 12-foot shadow. Find the measure of the
angle of elevation of the Sun to the nearest degree.
9.2
18°
x
A D
C
F
B
22
x
x2020
20
60
20x
2 12
x
6
4x
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
8
4√3
60°
30°
x
y
30 ft
x°
40 ft
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Chapter 8 62 Glencoe Geometry
Chapter 8 Test, Form 2C (continued)
12. A boat is 1000 meters from a cliff. If the angle of depression from the top of the cliff to the boat is 15°, how tall is the cliff? Round your answer to the nearest tenth.
13. A plane flying at an altitude of 10,000 feet begins descending when the end of the runway is 50,000 feet from a point on the ground directly below the plane. Find the measure of the angle of descent (depression) to the nearest degree.
14. Find x to the nearest tenth.
15. Find x to the nearest tenth.
16. A tree grew at a 3° slant from the vertical. At a point 50 feet from the tree, the angle of elevation to the top of the tree is 17°. Find the height of the tree to the nearest tenth of a foot.
17. Find x to the nearest degree.
18. In △XYZ, m∠X = 152, y = 15, and z = 19. Find x to the nearest tenth.
19. Find the magnitude and direction of the vector % AZ : A(8, 8)
and Z(1, −3).
20. Copy the vectors to find % s +
% t .
Bonus Find x.
11
75
x°
17° 93°
50 ft
x
x7
23°
146°
52°
26
37°x
1000 m
8
√6
5x
t*s*
12.
13.
14.
15.
16.
17.
18.
19.
20.
B:
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Chapter 8 63 Glencoe Geometry
Chapter 8 Test, Form 2D
1. Find the geometric mean between 3 √ " 6 and 5 √ " 6 .
For Questions 2–5, find x.
2. 3.
4. 5.
6. Find x.
7. In parallelogram ABCD, AD = 14
and m∠D = 60. Find AF.
8. Find x and y.
9. Find x to the nearest tenth.
10. An A-frame house is 45 feet high and
32 feet wide. Find the measure of the
angle that the roof makes with the
floor. Round to the nearest degree.
11. A 38-foot tree casts a 16-foot shadow. Find the measure of the
angle of elevation of the sun to the nearest degree.
16°
8.3
x
A D
C
F
B
30
x
x2424
24
80
30x
103
x
12
8
x
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
8
8√3
30°
60°
x
y
32 ft
x
45 ft
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Chapter 8 64 Glencoe Geometry
Chapter 8 Test, Form 2D (continued)
12. A boat is 2000 meters from a cliff. If the angle of depression from the top of the cliff to the boat is 10°, how tall is the cliff? Round your answer to the nearest tenth.
13. A plane flying at an altitude of 10,000 feet begins descending when the end of the runway is 60,000 feet from a point on the ground directly below the plane. Find the measure of the angle of descent (depression) to the nearest degree.
14. Find x to the nearest tenth.
15. Find x to the nearest degree.
16. A tree grew at a 3° slant from the vertical. At a point 60 feet from the tree, the angle of elevation to the top of the tree is 14°. Find the height of the tree to the nearest tenth of a foot.
17. Find x to the nearest degree.
18. In △XYZ, m∠X = 156, y = 18, and z = 21. Find x to the nearest tenth.
19. Find the magnitude and direction of the vector $ PQ : P(−2, 4)
and Q(−5, −6).
20. Copy the vectors to find $ d -
$ f .
Bonus Find x.
8 9
16
x°
14° 93°
60 ft
x
38°
17
6
x°
68°
42°
23x
2000 m
8
2√15
16
x
d* f*
12.
13.
14.
15.
16.
17.
18.
19.
20.
B:
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Chapter 8 65 Glencoe Geometry
Chapter 8 Test, Form 3
1. Find the geometric mean between 2−9
and 3−9
.
2. Find x in △PQR.
3. Find x in △XYZ.
4. If the length of one leg of a right triangle is three times the length of the other and the hypotenuse is 20, find the length of the shorter leg.
5. Find the measure of the altitude drawn to the hypotenuse of a right triangle with legs that measure 3 and 4.
6. Find x.
7. Richmond is 200 kilometers due east of Teratown and Hamilton is 150 kilometers directly north of Teratown. Find the shortest distance in kilometers between Hamilton and Richmond.
8. The measures of the sides of a triangle are 48, 55, and 73. Use the Pythagorean Theorem to classify the triangle as acute, obtuse, or right.
9. Find the exact perimeter of this square.
10. Find the exact perimeter of rectangle ABCD.
11. Find x and y.
12. △ABC is a 30°-60°-90° triangle with right angle A and
with −−AC as the longer leg. If A(-4, -2) and B(-4, 6), find the
coordinates of C.
15
30°
60°
xy
12
60°
A B
CD
17
8
9 3.5
x
65
2x
Q
P
R
S
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
8
x + 4 xX
WZ
Y
21√$
36√
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Chapter 8 66 Glencoe Geometry
Chapter 8 Test, Form 3 (continued)
13. If −−
AB ‖ −−−
CD , find x and the exact length of
−−− CD .
14. The angle of elevation from a point on the street to the top of a building is 29°. The angle of elevation from another point on the street, 50 feet farther away from the building, to the top of the building is 25°. To the nearest foot, how tall is the building?
15. The angle of depression from the top of a flagpole on top of a lighthouse to a boat on the ocean is 2.9°. The angle of depression from the bottom of the flagpole to the boat is 2.6°. If the boat is 400 feet away from shore and the lighthouse is right on the edge of the shore, how tall is the flagpole? Round your answer to the nearest foot.
16. In △JKL, m∠J = 26.8, m∠K = 19, and k = 17. Find ℓ to the nearest tenth.
17. Don hit a golf ball from the tee toward the hole which is 250 yards away. However, due to the wind, his drive was 5° off course. If the angle between the segment from the hole to the tee and the segment from the hole to the ball measures 97°, how far did Don drive the ball? Round to the nearest tenth of a yard.
18. In △HJK, h = 7, j = 12.3, and k = 7.9. Find m∠K. Round your answer to the nearest degree.
19. Find the magnitude and direction of the vector (WZ : W(15, 25)
and Z(10, −6).
20. Copy the vectors to find (a +
(b and
(a -
(b .
Bonus A 50-foot vertical pole that stands on a hillside. The slope of the hill side makes an angle of 10° with the horizontal. Two guy wires extend from the top of the pole to points on the hill 60 feet uphill and downhill from its base. Find the length of each guy wire to the nearest tenth of a foot.
12
10
60°45°x
A B
C D
8
13.
14.
15.
16.
17.
18.
19.
20.
B:
a*b*
Assessment
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Chapter 8 67 Glencoe Geometry
Chapter 8 Extended-Response Test
Demonstrate your knowledge by giving a clear, concise solution to
each problem. Be sure to include all relevant drawings and justify
your answers. You may show your solution in more than one way or
investigate beyond the requirements of the problem.
1. If the geometric mean between 10 and x is 6, what is x? Show how you
obtained your answer.
2.
a. Max used the following equations to find x in △PQR. Is Max correct?
Explain.
x = √ ## 8 · 2
x = √ ## 16
x = 4
b. If ∠PRQ is a right angle, what is the measure of −−−
PS ?
c. Is △PRS a 45°-45°-90° triangle? Explain.
3. To solve for x in a triangle, when would you use sin and when would you
use sin-1? Give an example for each type of situation.
4. Draw a diagram showing the angles of elevation and depression and label
each. How are the measures of these angles related?
5. Draw an obtuse triangle and label the vertices, the measures of two
angles, and the length of one side. Explain how to solve the triangle.
6. Irina is solving △ABC. She plans to first use the Law of Sines to find two
of the angles. Is Irina’s plan a good one? Explain.
4
15
12
A C
B
8 2
60°
x
P Q
R
S
8
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Chapter 8 68 Glencoe Geometry
Standardized Test Practice (Chapters 1–8)
1. If −− TA bisects ∠YTB,
−− TC bisects ∠BTZ, m∠YTA = 4y + 6, and m∠BTC = 7y - 4, find m∠CTZ. (Lesson 1-4)
A 52 B 38 C 25 D 8
2. Which statement is always true? (Lesson 2-5)
F If right triangle QPR has sides q, p, and r, where r is the hypotenuse,
then r2 = p2 + q2.
G If −− EF ‖
−− HJ , then EF = HJ.
H If −− KL and
−− VT are cut by a transversal, then −− KL ‖ −− VT .
J If −−− DR and
−−− RH are congruent, then R bisects −−− DH .
3. The equation for $ %& PT is y - 2 = 8(x + 3). Determine an equation for a
line perpendicular to $ %& PT . (Lesson 3-4)
A y = 1 − 8 x - 7 B y = 8x - 13 C y = - 1 −
8 x + 2 D y = -8x
4. Angle Y in △XYZ measures 90°. −− XY and
−− YZ each measure 16 meters. Classify △XYZ. (Lesson 4-1)
F acute and isosceles H right and scalene
G equiangular and equilateral J right and isosceles
5. Two sides of a triangle measure 4 inches and 9 inches. Determine which cannot be the perimeter of the triangle. (Lesson 5-3)
A 19 in. B 21 in. C 23 in. D 26 in.
6. △ABC ∼ △STR, so AB −
CA = ? . (Lesson 6-2)
F AB
− BC
G ST
− RS
H TR
− RS
J RS
− ST
7. The Petronas Towers in Kuala Lumpur, Malaysia, are 452 meters tall. A woman who is 1.75 meters tall stands 120 meters from the base of one tower. Find the angle of elevation between the woman’s hat and the top of the tower. Round to the nearest tenth. (Lesson 8-5)
A 14.8° B 34.9° C 55° D 75.1°
8. Which equation can be used to find x? (Lesson 8-4)
F x = y sin 73° H x = y −
cos 73°
G x = y cos 73° J x = y −
sin 73°
9. Which inequality describes the possible values
of x? (Lesson 5-6)
A x < 4 C x > 2B x < 2 D x > 4
85°117°
1212
14
5x - 6
73°
x
y
Y ZT
A CB
1. A B C D
2. F G H J
3. A B C D
4. F G H J
5. A B C D
6. F G H J
7. A B C D
8. F G H J
9. A B C D
8
Part 1: Multiple Choice
Instructions: Fill in the appropriate circle for the best answer.
Assessment
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Chapter 8 69 Glencoe Geometry
15.
16.
Standardized Test Practice (continued)
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
10. Ashley wants to make a poster for her campaign from a photograph. She uses a photocopier to enlarge the 4 inch by 6 inch photograph. What are the dimensions of the poster if she increases the size of the photograph by a scale factor of 5? (Lesson 7-7)
F 2000 inches by 3000 inches H 9 inches by 11 inches
G 0.8 inches by 1.2 inches J 20 inches by 30 inches
11. Find x. (Lesson 6-2)
A -3 C 5
B 3 D 9
12. Trapezoid ABCD has vertices A(1, 6), B(-2, 6), C(-10, -10), and
D(20, -10). Find the measure of ABCD’s midsegment to the nearest tenth. (Lesson 6-6)
F 3 G 5.3 H 7.2 J 16.5
For Questions 13 and 14, use the
figure to the right.
13. Find QP to the nearest tenth. (Lesson 8-2)
A 7.5 B 12 C 18.3 D 19.6
14. Find LM. (Lesson 8-3)
F 5 G 5 √ " 3 H 9 J 10 √ " 3
60°30°
10
9
P
M
L
Q
R
5
10. F G H J
11. A B C D
12. F G H J
13. A B C D
14. F G H J
15. Find x so that ℓ ‖ m. (Lesson 3-5)
16. Find c to the nearest tenth. (Lesson 8-6)
(134 - 4x)°(6x + 17)°
mℓ
8
Part 2: Gridded Response
Instructions: Enter your answer by writing each digit of the answer in a column box and then
shading in the appropriate circle that corresponds to that entry.
26.1°
27.7°
126.2°
19
33
A
B
C
c
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Chapter 8 70 Glencoe Geometry
Standardized Test Practice (continued)
For Questions 17 and 18, complete
the following proof. (Lesson 2-7)
Given: −−
JK " −−−
LM
−−−
HJ " −−
KL
Prove: −−−
HK " −−−
KM
Proof:
Statements Reasons
1. −−
JK " −−−
LM , −−−
HJ " −−
KL 1. Given
2. JK = LM, HJ = KL 2. (Question 17)
3. (Question 18) 3. Segment Addition Post.
4. HJ + JK = KL + LM 4. Add. Prop. of Equality
5. HK = KM 5. Substitution Prop.
6. −−−
HK " −−−
KM 6. Def. of " segments
For Questions 19 and 20, use the figure
at the right.
19. Find the measure of the numbered angles if m∠ABC = 57 and m∠BCE = 98. (Lesson 4-2)
20. If −−−
BD is a midsegment of △ABC, AD = 2x - 6, and DC = 22.5 - 4x, find AC. (Lesson 5-2)
21. If △DEF " △HJK, m∠D = 26, m∠J = 3x + 5, and m∠F = 92, find x. (Lesson 4-3)
22. Use the Exterior Angle Inequality Theorem to list all of the angles with measures that are less than m∠1. (Lesson 5-3)
23. a. Determine whether △EFH ∼ △JGH. Justify your answer. (Lesson 7-3)
b. If G is the midpoint of −−−
FH , find x. (Lesson 8-3)
c. Use the value of x you found in part b to find the scale factor of △EFH to △JGH. (Lesson 7-3)
1 2 4
3 5
6 7
8A DC
B
E
CB
A
D
25°1
2
34
5
41°
H
J
K L M
17.
18.
19.
20.
21.
22.
23a.
23b.
23c.
Part 3: Short Response
Instructions: Write your answer in the space provided.
60°18
30°E JH
F
Gx
8
Answers
Co
pyri
gh
t ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f T
he
Mc
Gra
w-H
ill C
om
pa
nie
s,
Inc
.
Chapter 8 A1 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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ME
DA
TE
PE
RIO
D
Chapter Resources
Ch
ap
ter
8
3
Gle
nc
oe
Ge
om
etr
y
B
efo
re y
ou
beg
in C
ha
pte
r 8
•
R
ead
each
sta
tem
en
t.
•
D
eci
de w
heth
er
you
Agre
e (
A)
or
Dis
agre
e (
D)
wit
h t
he s
tate
men
t.
•
W
rite
A o
r D
in
th
e f
irst
colu
mn
OR
if
you
are
not
sure
wh
eth
er
you
agre
e o
r d
isagre
e,
wri
te N
S (
Not
Su
re).
An
tici
pati
on
Gu
ide
Rig
ht
Tri
an
gle
s a
nd
Tri
go
no
metr
y
A
fter y
ou
co
mp
lete
Ch
ap
ter 8
•
Rere
ad
each
sta
tem
en
t an
d c
om
ple
te t
he l
ast
colu
mn
by e
nte
rin
g a
n A
or
a D
.
•
Did
an
y o
f you
r op
inio
ns
abou
t th
e s
tate
men
ts c
han
ge f
rom
th
e f
irst
colu
mn
?
•
For
those
sta
tem
en
ts t
hat
you
mark
wit
h a
D,
use
a p
iece
of
pap
er
to w
rite
an
exam
ple
of
wh
y y
ou
dis
agre
e.
8 Ste
p 1
ST
EP
1A
, D
, o
r N
SS
tate
men
tS
TE
P 2
A o
r D
1
. T
he g
eom
etr
ic m
ean
betw
een
tw
o n
um
bers
is
the
posi
tive s
qu
are
root
of
their
pro
du
ct.
A
2
. A
n a
ltit
ud
e d
raw
n f
rom
th
e r
igh
t an
gle
of
a r
igh
t tr
ian
gle
to i
ts h
yp
ote
nu
se s
ep
ara
tes
the t
rian
gle
in
to
two c
on
gru
en
t tr
ian
gle
s.D
3
. In
a r
igh
t tr
ian
gle
, th
e l
en
gth
of
the h
yp
ote
nu
se i
s equ
al
to t
he s
um
of
the l
en
gth
s of
the l
egs.
D
4
. If
an
y t
rian
gle
has
sid
es
wit
h l
en
gth
s 3,
4,
an
d 5
, th
en
th
at
tria
ngle
is
a r
igh
t tr
ian
gle
.A
5
. If
th
e t
wo a
cute
an
gle
s of
a r
igh
t tr
ian
gle
are
45°,
then
th
e l
en
gth
of
the h
yp
ote
nu
se i
s √
"
2 t
imes
the
len
gth
of
eit
her
leg.
A
6
. In
an
y t
rian
gle
wh
ose
an
gle
measu
res
are
30°,
60°,
an
d 9
0°,
th
e h
yp
ote
nu
se i
s √
"
3 t
imes
as
lon
g a
s th
e
short
er
leg.
D
7
. T
he s
ine r
ati
o o
f an
an
gle
of
a r
igh
t tr
ian
gle
is
equ
al
to t
he l
en
gth
of
the a
dja
cen
t si
de d
ivid
ed
by t
he
len
gth
of
the h
yp
ote
nu
se.
D
8
. T
he t
an
gen
t of
an
an
gle
of
a r
igh
t tr
ian
gle
wh
ose
si
des
have l
en
gth
s 3,
4,
an
d 5
wil
l be s
mall
er
than
th
e
tan
gen
t of
an
an
gle
of
a r
igh
t tr
ian
gle
wh
ose
sid
es
have l
en
gth
s 6,
8,
an
d 1
0.
D
9
. T
rigon
om
etr
ic r
ati
os
can
be u
sed
to s
olv
e p
roble
ms
involv
ing a
ngle
s of
ele
vati
on
an
d a
ngle
s of
dep
ress
ion
.A
10
. T
he L
aw
of
Sin
es
can
on
ly b
e u
sed
in
rig
ht
tria
ngle
s.D
Ste
p 2
Lesson 8-1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
5
Gle
nc
oe
Ge
om
etr
y
Stu
dy
Gu
ide a
nd
In
terv
en
tio
n
Geo
metr
ic M
ean
Ge
om
etr
ic M
ea
n
Th
e g
eo
me
tric
me
an
betw
een
tw
o n
um
bers
is
the p
osi
tive s
qu
are
ro
ot
of
their
pro
du
ct.
For
two p
osi
tive n
um
bers
a a
nd
b,
the g
eom
etr
ic m
ean
of
a a
nd
b i
s
the p
osi
tive n
um
ber
x i
n t
he p
rop
ort
ion
a
−
x =
x −
b .
Cro
ss m
ult
iply
ing g
ives
x2 =
ab,
so x
= √
""
ab .
F
ind
th
e g
eo
me
tric
me
an
be
twe
en
ea
ch
pa
ir o
f n
um
be
rs.
a.
12
an
d 3
x =
√ "
"
ab
Definitio
n o
f geom
etr
ic m
ean
=
√ "
""
12 . 3
a
= 1
2 a
nd b
= 3
=
√ "
""
""
(2 .
2 .
3)
. 3
Facto
r.
=
6
Sim
plif
y.
Th
e g
eom
etr
ic m
ean
betw
een
12 a
nd
3 i
s 6.
b.
8 a
nd
4
x =
√ "
"
ab
Definitio
n o
f geom
etr
ic m
ean
=
√ "
"
8 . 4
a
= 8
and b
= 4
=
√ "
""
"
(2 .
4)
. 4
Facto
r.
=
√ "
""
16 .
2
Associa
tive P
ropert
y
=
4 √
"
2
Sim
plif
y.
Th
e g
eom
etr
ic m
ean
betw
een
8 a
nd
4
is 4
√ "
2 o
r abou
t 5.7
.
Exerc
ises
Fin
d t
he
ge
om
etr
ic m
ea
n b
etw
ee
n e
ach
pa
ir o
f n
um
be
rs.
1. 4 a
nd
4
2. 4 a
nd
6
3. 6 a
nd
9
4. 1
−
2 a
nd
2
5. 12 a
nd
20
6. 4 a
nd
25
7. 16 a
nd
30
8. 10 a
nd
100
9. 1
−
2 a
nd
1
−
4
10. 17 a
nd
3
11. 4 a
nd
16
12. 3 a
nd
24
8-1 Exam
ple
4
√ "
"
24 o
r 2 √
"
6 ≈
4.9
√ "
"
240 o
r 4 √
""
15 ≈
15.5
√ "
""
1000 o
r 10 √
""
10 ≈
31.6
10
√"
"54
or
3√
" 6≈
7.3
1
√ "
"
480 o
r 4 √
""
30 ≈
21.9
√ "
"
51 ≈
7.1
8
√ "
"
72 o
r 6 √
"
2 ≈
8.5
√ "
1
−
8 o
r √
"
2
−
4 ≈
0.4
Answers (Anticipation Guide and Lesson 8-1)
Co
pyrig
ht ©
Gle
nc
oe
/Mc
Gra
w-H
ill, a d
ivis
ion
of T
he
Mc
Gra
w-H
ill Co
mp
an
ies, In
c.
Chapter 8 A2 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
6
Gle
nc
oe
Ge
om
etr
y
z
y
x
15
R QP
S25
Stu
dy
Gu
ide a
nd
In
terv
en
tio
n
(con
tin
ued
)
Geo
metr
ic M
ean
Ge
om
etr
ic M
ea
ns
in R
igh
t T
ria
ng
les
In t
he d
iagra
m,
△ A
BC
∼ △
AD
B ∼
△ B
DC
. A
n a
ltit
ud
e t
o t
he h
yp
ote
nu
se o
f a r
igh
t tr
ian
gle
form
s tw
o r
igh
t tr
ian
gle
s. T
he t
wo t
rian
gle
s are
sim
ilar
an
d
each
is
sim
ilar
to t
he o
rigin
al
tria
ngle
.
U
se
rig
ht
△ A
BC
wit
h
−−
BD
⊥ −
−
AC
. D
escrib
e t
wo
ge
om
etr
ic
me
an
s.
a.
△ A
DB
∼ △
BD
C s
o A
D
−
BD
= B
D
−
CD
.
In △
AB
C,
the a
ltit
ud
e i
s th
e g
eom
etr
ic
mean
betw
een
th
e t
wo s
egm
en
ts o
f th
e
hyp
ote
nu
se.
b.
△ A
BC
∼ △
AD
B a
nd △
AB
C ∼
△ B
DC
,
so
AC
−
AB
= A
B
−
AD
an
d A
C
−
BC
= B
C
−
DC
.
In △
AB
C,
each
leg i
s th
e g
eom
etr
ic
mean
betw
een
th
e h
yp
ote
nu
se a
nd
th
e
segm
en
t of
the h
yp
ote
nu
se a
dja
cen
t to
that
leg.
F
ind
x,
y,
an
d z
.
15 =
√ &
&&
&
RP
' S
P
Geom
etr
ic M
ean (
Leg)
Theore
m
15 =
√ &
&
25x
RP
= 2
5 a
nd S
P =
x
225 =
25x
Square
each s
ide.
9 =
x
Div
ide e
ach s
ide b
y 2
5.
Th
en
y =
RP
– S
P
=
25 –
9
=
16
z =
√ &
&&
&
RS
' R
P
Geom
etr
ic M
ean (
Leg)
Theore
m
=
√ &
&&
16 '
25
RS
= 1
6 a
nd R
P =
25
=
√ &
&
400
Multip
ly.
=
20
Sim
plif
y.
Exerc
ises
Fin
d x
, y,
an
d z
to
th
e n
ea
re
st
ten
th.
1.
x
13
2.
z
xy
5
2
3.
zx
y 81
x =
√ %
3 ≈
1.7
x =
√ %
%
10 ≈
3.2
; x =
3;
y =
√ %
%
14 ≈
3.7
; y =
√ %
%
72 o
r 6 √
%
2 ≈
8.5
;
z =
√ %
%
35 ≈
5.9
z =
√ %
8 o
r 2 √
%
2 ≈
2.8
4.
xy
1
√3
√"12
5.
x
zy
2
2
6.
xz
y
62
x =
2;
x =
2;
x =
√ %
%
12 o
r 2 √
%
3 ≈
3.5
;
y =
3
y =
√ %
8 o
r 2 √
%
2 ≈
2.8
; y =
√ %
8 o
r 2 √
%
2 ≈
2.8
;
z =
√ %
8 o
r 2 √
%
2 ≈
2.8
z =
√ %
%
24 o
r 2 √
%
6 ≈
4.9
8-1
CDB
A
Exam
ple
1Exam
ple
2
Lesson 8-1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
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PE
RIO
D
Ch
ap
ter
8
7
Gle
nc
oe
Ge
om
etr
y
Sk
ills
Pra
ctic
e
Geo
metr
ic M
ean
Fin
d t
he
ge
om
etr
ic m
ea
n b
etw
ee
n e
ach
pa
ir o
f n
um
be
rs.
1. 2 a
nd
8
2. 9 a
nd
36
3. 4 a
nd
7
4. 5 a
nd
10
5. 28 a
nd
14
6. 7 a
nd
36
Writ
e a
sim
ila
rit
y s
tate
me
nt
ide
nti
fyin
g t
he
th
re
e s
imil
ar t
ria
ng
les i
n t
he
fig
ure
.
7.
C
D
B
A
8.
L
M N
P
9.
GEH
F
10.
RT
S U
Fin
d x
, y a
nd
z.
11.
39y
x
z
12.
10
4
y
x
z
13.
15
4
y
xz
14.
25y
x
z
8-1
4
18
√
%%
28 o
r 2 √
%
7 ≈
5.3
√ %
%
50 o
r 5 √
%
2 ≈
7.1
√ %
%
392 o
r 14 √
%
2 ≈
19.8
√
%%
252 o
r 6 √
%
7 ≈
15.9
△A
CB
∼ △
CD
B ∼
△A
DC
△M
NL ∼
△N
PL ∼
△M
PN
△E
GF ∼
△G
HF ∼
△E
HG
△R
ST ∼
△S
UT ∼
△R
US
6;
√ %
%
108 o
r 6 √
%
3 ≈
10.4
;
√ %
%
40 o
r 2 √
%%
10 ≈
6.3
; √
%%
56 o
r 2 √
%%
14 ≈
7.5
;
√ %
%
27 o
r 3 √
%
3 ≈
5.2
√
%%
140 o
r 2 √
%%
35 ≈
11.8
√%
%60
or
2√
%%
15
≈ 7
.7;
12.5
; √
%%
29
≈ 5
.4;
√%
%%
181.2
5≈
13.5
√ %
%
285 o
r 2 √
%%
19 ≈
16.9
;
√ %
%
76 ≈
8.7
Answers (Lesson 8-1)
Answers
Co
pyri
gh
t ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f T
he
Mc
Gra
w-H
ill C
om
pa
nie
s,
Inc
.
Chapter 8 A3 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
8
Gle
nc
oe
Ge
om
etr
y
Practi
ce
Geo
metr
ic M
ean
Fin
d t
he
ge
om
etr
ic m
ea
n b
etw
ee
n e
ach
pa
ir o
f n
um
be
rs.
1. 8 a
nd
12
2. 3 a
nd
15
3.
4
−
5 a
nd
2
Writ
e a
sim
ila
rit
y s
tate
me
nt
ide
nti
fyin
g t
he
th
re
e s
imil
ar t
ria
ng
les i
n t
he
fig
ure
.
4.
U
TA
V
5.
KLJ
M
Fin
d x
, y,
an
d z
.
6.
23
z
xy
8
7.
zx
y625
8.
x
y
2
3
z
9.
xy1
0z
20
10
. C
IVIL
A
n a
irp
ort
, a f
act
ory
, an
d a
sh
op
pin
g c
en
ter
are
at
the v
ert
ices
of
a r
igh
t tr
ian
gle
form
ed
by t
hre
e h
igh
ways.
Th
e a
irp
ort
an
d f
act
ory
are
6.0
mil
es
ap
art
. T
heir
dis
tan
ces
from
th
e s
hop
pin
g c
en
ter
are
3.6
mil
es
an
d 4
.8 m
iles,
resp
ect
ively
. A
serv
ice r
oad
wil
l be
con
stru
cted
fro
m t
he s
hop
pin
g c
en
ter
to t
he h
igh
way t
hat
con
nect
s th
e a
irp
ort
an
d
fact
ory
. W
hat
is t
he s
hort
est
poss
ible
len
gth
for
the s
erv
ice r
oad
? R
ou
nd
to t
he n
eare
st
hu
nd
red
th.
8-1
√ "
"
96 o
r 4 √
"
6 ≈
9.8
√ "
"
45 o
r 3 √
"
5 ≈
6.7
√ "
8
−
5 o
r 2
√ "
"
10
−
5
≈ 1
.3
△
VU
T ∼
△U
AT ∼
△V
AU
△
JLK
∼ △
LM
K ∼
△JM
L
x
= √
""
184 o
r 2 √
""
46 ≈
13.6
; x
= √
""
114 ≈
10.7
;
y=
√"
"248
or
2√
""
62
≈ 1
5.7
y=
√"
"150
or
5√
" 6≈
12.2
z
= √
""
713 ≈
26.7
z
= √
""
475 o
r 5 √
""
19 ≈
21.8
x
= 4
.5;
x
= 1
5;
x
= √
""
13 ≈
3.6
; 6.5
y =
5;
z =
√ "
"
300 o
r 10 √
"
3 ≈
17.3
2.8
8 m
i
Lesson 8-1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
9
Gle
nc
oe
Ge
om
etr
y
1. S
QU
AR
ES
W
ilm
a h
as
a r
ect
an
gle
of
dim
en
sion
s ℓ b
y w
. S
he w
ou
ld l
ike t
o
rep
lace
it
wit
h a
squ
are
th
at
has
the
sam
e a
rea.
Wh
at
is t
he s
ide l
en
gth
of
the s
qu
are
wit
h t
he s
am
e a
rea a
s
Wil
ma’s
rect
an
gle
?
2. E
QU
ALIT
Y G
retc
hen
com
pu
ted
th
e
geom
etr
ic m
ean
of
two n
um
bers
. O
ne o
f
the n
um
bers
was
7 a
nd
th
e g
eom
etr
ic
mean
tu
rned
ou
t to
be 7
as
well
. W
hat
was
the o
ther
nu
mber?
3. V
IEW
ING
AN
GLE
A
ph
oto
gra
ph
er
wan
ts t
o t
ak
e a
pic
ture
of
a b
each
fro
nt.
His
cam
era
has
a v
iew
ing a
ngle
of
90°
an
d h
e w
an
ts t
o m
ak
e s
ure
tw
o p
alm
trees
loca
ted
at
poin
ts A
an
d B
in
th
e
figu
re a
re j
ust
in
sid
e t
he e
dges
of
the
ph
oto
gra
ph
.
Walkway
90 f
t4
0 f
tA
B
x
AB
x
He w
alk
s ou
t on
a w
alk
way t
hat
goes
over
the o
cean
to g
et
the s
hot.
If
his
cam
era
has
a v
iew
ing a
ngle
of
90°,
at
wh
at
dis
tan
ce d
ow
n t
he w
alk
way s
hou
ld
he s
top
to t
ak
e h
is p
hoto
gra
ph
?
4
. E
XH
IBIT
ION
S A
mu
seu
m h
as
a f
am
ou
s
statu
e o
n d
isp
lay.
Th
e c
ura
tor
pla
ces
the
statu
e i
n t
he c
orn
er
of
a r
ect
an
gu
lar
room
an
d b
uil
ds
a 1
5-f
oot-
lon
g r
ail
ing i
n
fron
t of
the s
tatu
e.
Use
th
e i
nfo
rmati
on
belo
w t
o f
ind
how
clo
se v
isit
ors
wil
l be
able
to g
et
to t
he s
tatu
e.
Sta
tue
12
ft
15 ft
9 f
tx
5
. C
LIF
FS
A
bri
dge c
on
nect
s to
a t
un
nel
as
show
n i
n t
he f
igu
re.
Th
e b
rid
ge i
s
180 f
eet
above t
he g
rou
nd
. A
t a d
ista
nce
of
235 f
eet
alo
ng t
he b
rid
ge o
ut
of
the
tun
nel,
th
e a
ngle
to t
he b
ase
an
d
sum
mit
of
the c
liff
is
a r
igh
t an
gle
.
Clif
f
23
5 f
t
180 ft
dx
h
a.
Wh
at
is t
he h
eig
ht
of
the c
liff
? R
ou
nd
to t
he n
eare
st w
hole
nu
mber.
b.
How
hig
h i
s th
e c
liff
fro
m b
ase
to
sum
mit
? R
ou
nd
to t
he n
eare
st w
hole
nu
mber.
c.
Wh
at
is t
he v
alu
e o
f d
? R
ou
nd
to t
he
neare
st w
hole
nu
mber.
Wo
rd
Pro
ble
m P
racti
ce
Geo
metr
ic M
ean
8-1
7
.2 f
t
6
0 f
t
387 f
t
√ "
"
ℓw
, th
e g
eo
metr
ic m
ean
of
ℓ
an
d w
7
4
87 f
t
3
07 f
t
Answers (Lesson 8-1)
Co
pyrig
ht ©
Gle
nc
oe
/Mc
Gra
w-H
ill, a d
ivis
ion
of T
he
Mc
Gra
w-H
ill Co
mp
an
ies, In
c.
Chapter 8 A4 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
10
G
len
co
e G
eo
me
try
Math
em
ati
cs a
nd
Mu
sic
Pyth
agora
s, a
Gre
ek
ph
iloso
ph
er
wh
o l
ived
du
rin
g t
he s
ixth
cen
tury
B.C
.,
beli
eved
th
at
all
natu
re,
beau
ty,
an
d h
arm
on
y c
ou
ld b
e e
xp
ress
ed
by w
hole
-n
um
ber
rela
tion
ship
s. M
ost
peop
le r
em
em
ber
Pyth
agora
s fo
r h
is t
each
ings
abou
t ri
gh
t tr
ian
gle
s. (
Th
e s
um
of
the s
qu
are
s of
the l
egs
equ
als
th
e s
qu
are
of
the h
yp
ote
nu
se.)
Bu
t P
yth
agora
s als
o d
isco
vere
d r
ela
tion
ship
s betw
een
th
e
mu
sica
l n
ote
s of
a s
cale
. T
hese
rela
tion
ship
s ca
n b
e e
xp
ress
ed
as
rati
os.
C
D
E
F
G
A
B
C
′1
−
1
8
−
9
4
−
5
3
−
4
2
−
3
3
−
5
8
−
15
1
−
2
Wh
en
you
pla
y a
str
inged
in
stru
men
t,
Th
e C
str
ing c
an
be u
sed
you
pro
du
ce d
iffe
ren
t n
ote
s by p
laci
ng
to p
rod
uce
F b
y p
laci
ng
you
r fi
nger
on
dif
fere
nt
pla
ces
on
a s
trin
g.
a f
inger
3
−
4 o
f th
e w
ay
Th
is i
s th
e r
esu
lt o
f ch
an
gin
g t
he l
en
gth
alo
ng t
he s
trin
g.
of
the v
ibra
tin
g p
art
of
the s
trin
g.
Su
pp
ose
a C
str
ing
ha
s a
le
ng
th o
f 1
6 i
nch
es.
Writ
e a
nd
so
lve
pro
po
rti
on
s t
o d
ete
rm
ine
wh
at
len
gth
of
str
ing
wo
uld
ha
ve
to
vib
ra
te t
o p
ro
du
ce
th
e r
em
ain
ing
no
tes o
f th
e s
ca
le.
1
. D
2. E
3. F
4
. G
5. A
6. B
7. C
′
8
. C
om
ple
te t
o s
how
th
e d
ista
nce
betw
een
fin
ger
posi
tion
s on
th
e 1
6-i
nch
C
str
ing f
or
each
note
. F
or
exam
ple
, C
(16)
- D
(14 2
−
9 ) =
1 7
−
9 .
C
D
E
F
G
A
B
C
′
3 4of
C s
trin
g
En
rich
men
t
1 7 − 9
in
.
8-1
3 1 − 5
in
.4 i
n.
5 1 − 3
in
.6 2 − 5
in
.7 7
−
15 in
.8 i
n.
10 2
−
3 in
.8
8
−
15 in
.
12 i
n.
8 i
n.
14 2 − 9
in
.12 4 − 5
in
.
9 3
−
5 in
.
Lesson 8-2
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
11
G
len
co
e G
eo
me
try
Stu
dy
Gu
ide a
nd
In
terv
en
tio
n
Th
e P
yth
ag
ore
an
Th
eo
rem
an
d I
ts C
on
vers
e
Th
e P
yth
ag
ore
an
Th
eo
rem
In
a r
igh
t tr
ian
gle
, th
e s
um
of
the
squ
are
s of
the l
en
gth
s of
the l
egs
equ
als
th
e s
qu
are
of
the l
en
gth
of
the h
yp
ote
nu
se.
If t
he t
hre
e w
hole
nu
mbers
a,
b,
an
d c
sati
sfy t
he e
qu
ati
on
a
2 +
b2 =
c2,
then
th
e n
um
bers
a,
b,
an
d c
form
a
Py
tha
go
re
an
trip
le.
ca
bA
CB
a.
Fin
d a
.
a
12
13
ACB
a
2 +
b2 =
c2
Pyth
agore
an T
heore
m
a
2 +
12
2 =
13
2
b =
12,
c =
13
a
2 +
144 =
169
Sim
plif
y.
a
2 =
25
Subtr
act.
a
= 5
T
ake t
he p
ositiv
e s
quare
root
of
each s
ide.
b.
Fin
d c
. c
30
20
ACB
a
2 +
b2 =
c2
Pyth
agore
an T
heore
m
20
2 +
30
2 =
c2
a =
20,
b =
30
400 +
900 =
c2
Sim
plif
y.
1300 =
c2
Add.
√
$$
1300 =
c
Take t
he p
ositiv
e s
quare
root
of
each s
ide.
36.1
≈ c
U
se a
calc
ula
tor.
Fin
d x
.
1.
x
33
2.
x 15
9
3.
x65
25
4.
x
5 9
4 9
5.
x
33
16
6.
x
11
28
Use
a P
yth
ag
ore
an
Trip
le t
o f
ind
x.
7
. 8
17
x
8
. 45
24
x
9
. 28
96
x
8-2 Exam
ple
Exerc
ises
△ A
BC
is a
rig
ht
tria
ngle
.
so a
2 +
b2 =
c2.
1 − 3√
##
#1345
≈ 3
6.7
√#
#663
≈ 2
5.7
√ #
#
18 o
r 3
√ #
2 ≈
4.2
12
60
15
51
100
Answers (Lesson 8-1 and Lesson 8-2)
Answers
Co
pyri
gh
t ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f T
he
Mc
Gra
w-H
ill C
om
pa
nie
s,
Inc
.
Chapter 8 A5 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
12
G
len
co
e G
eo
me
try
c
ab
A
C
B
Stu
dy
Gu
ide a
nd
In
terv
en
tio
n
(con
tin
ued
)
Th
e P
yth
ag
ore
an
Th
eo
rem
an
d I
ts C
on
vers
e
Co
nv
ers
e o
f th
e P
yth
ag
ore
an
Th
eo
rem
If
th
e s
um
of
the s
qu
are
s
of
the l
en
gth
s of
the t
wo s
hort
er
sid
es
of
a t
rian
gle
equ
als
th
e s
qu
are
of
the l
en
gth
s of
the l
on
gest
sid
e,
then
th
e t
rian
gle
is
a r
igh
t tr
ian
gle
.
You
can
als
o u
se t
he l
en
gth
s of
sid
es
to c
lass
ify a
tri
an
gle
. If a
2 +
b2 =
c2,
then
if
a2
+ b
2 =
c2 t
hen
△A
BC
is
a r
igh
t tr
ian
gle
. △
AB
C is a
rig
ht
tria
ngle
.
if
a2
+ b
2 >
c2 t
hen
△A
BC
is
acu
te.
if
a2
+ b
2 <
c2 t
hen
△A
BC
is
obtu
se.
D
ete
rm
ine
wh
eth
er △
PQ
R i
s a
rig
ht
tria
ng
le.
a
2 +
b2 "
c2
Com
pare
c2
and a
2 +
b2
10
2 +
(10 √
$
3 )2
" 2
02
a =
10,
b =
10 √
$
3 ,
c =
20
100 +
300 "
400
Sim
plif
y.
400 =
400
%
Add.
Sin
ce c
2 =
an
d a
2 +
b2,
the t
rian
gle
is
a r
igh
t tr
ian
gle
.
Exerc
ises
De
term
ine
wh
eth
er e
ach
se
t o
f m
ea
su
re
s c
an
be
th
e m
ea
su
re
s o
f th
e s
ide
s o
f a
tria
ng
le.
If s
o,
cla
ssif
y t
he
tria
ng
le a
s a
cu
te,
ob
tuse,
or r
igh
t. J
usti
fy y
ou
r a
nsw
er.
1. 30,
40,
50
2. 20,
30,
40
3. 18,
24,
30
4. 6,
8,
9
5.
6,
12,
18
6. 10,
15,
20
7. √
$
5 ,
√ $
$
12 ,
√ $
$
13
8.
2,
√ $
8 ,
√ $
$
12
9. 9,
40,
41
20
10
10√3
8-2 Exam
ple
y
es,
rig
ht;
yes,
ob
tuse;
yes,
rig
ht;
50
2 =
30
2 +
40
2
40
2 >
20
2 +
30
2
30
2 =
24
2 +
18
2
y
es,
acu
te;
n
o;
6 +
12 =
18
yes,
ob
tuse;
92 <
62 +
82
20
2 >
10
2 +
15
2
y
es,
acu
te;
yes,
rig
ht;
y
es,
rig
ht;
(
√ #
#
13 ) 2
< ( √
#
5 ) 2
+ (
√ #
#
12 ) 2
( √ #
#
12 ) 2
= ( √
#
8 ) 2
+ 2
2
41
2 =
40
2 +
92
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 8-2
Ch
ap
ter
8
13
G
len
co
e G
eo
me
try
Sk
ills
Pra
ctic
e
Th
e P
yth
ag
ore
an
Th
eo
rem
an
d I
ts C
on
vers
e
Fin
d x
.
1.
x 12
9
2.
x
1213
3.
x
12
32
4.
x12.5
25
5.
x9
9
8
6.
x31
14
Use
a P
yth
ag
ore
an
Trip
le t
o f
ind
x.
7.
5
12x
8.
8
10
x
9.
20
12
x
10
.
65
25
x
11
. 1
2.
5048
x
De
term
ine
wh
eth
er e
ach
se
t o
f n
um
be
rs c
an
be
me
asu
re
of
the
sid
es o
f a
tria
ng
le.
If s
o,
cla
ssif
y t
he
tria
ng
le a
s a
cu
te,
ob
tuse,
or r
igh
t. J
usti
fy y
ou
r a
nsw
er.
13
. 7,
24,
25
14
. 8,
14,
20
15
. 12.5
, 13,
26
16
. 3 √
"
2 ,
√ "
7 ,
4
17
. 20,
21,
29
18
. 32,
35,
70
8-2
40
24
x
1
5
5
√
##
#
1168 ≈
34.2
√#
##
468.7
5≈
21.7
√#
#65
≈ 8
.1√
##
#1157
≈ 3
4.0
Y
es,
rig
ht
tria
ng
le
Yes,
ob
tuse t
rian
gle
N
o,
12.5
+ 1
3 <
26
7 2 +
24 2 =
25 2
8 2 +
14 2 <
20 2
1
3
6
16
6
0
32
14
Y
es,
acu
te t
rian
gle
Yes,
rig
ht
tria
ng
le
No
, 32 +
35 <
70
( 3 √
#
2 ) 2
+ √
#
7 2 >
42
20 2 +
21 2
= 2
9 2
Answers (Lesson 8-2)
Co
pyrig
ht ©
Gle
nc
oe
/Mc
Gra
w-H
ill, a d
ivis
ion
of T
he
Mc
Gra
w-H
ill Co
mp
an
ies, In
c.
Chapter 8 A6 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
14
G
len
co
e G
eo
me
try
Practi
ce
Th
e P
yth
ag
ore
an
Th
eo
rem
an
d I
ts C
on
vers
e
Fin
d x
.
1.
x
13
23
2.
x
34
21
3.
x
26
26
18
4.
x
34
22
5.
x
16
14
6.
x
24
24
42
Use
a P
yth
ag
ore
an
Trip
le t
o f
ind
x.
7.
36
27
x
8.
120
136
x
9.
65
39
x
10.
42
15
0
x
De
term
ine
wh
eth
er e
ach
se
t o
f n
um
be
rs c
an
be
me
asu
re
of
the
sid
es o
f a
tria
ng
le.
If s
o,
cla
ssif
y t
he
tria
ng
le a
s a
cu
te,
ob
tuse,
or r
igh
t. J
usti
fy y
ou
r a
nsw
er.
11. 10,
11,
20
12. 12,
14,
49
13. 5 √
"
2 ,
10,
11
14. 21.5
, 24,
55.5
15. 30,
40,
50
16. 65,
72,
97
17
. C
ON
ST
RU
CT
ION
T
he b
ott
om
en
d o
f a r
am
p a
t a w
are
hou
se i
s 10 f
eet
from
th
e b
ase
of
the m
ain
dock
an
d i
s 11 f
eet
lon
g.
How
h
igh
is
the d
ock
?
11 f
t?
dock
ram
p
10 f
t
8-2
√ "
"
698 ≈
26.4
√ "
"
715 ≈
26.7
√ "
"
595 ≈
24.4
√ "
""
1640 ≈
40.5
√ "
"
60 ≈
7.7
√ "
"
135 ≈
11.6
ab
ou
t 4.6
ft
hig
h
4
5
5
2
y
es,
ob
tuse t
rian
gle
; n
o,
12 +
14
< 4
9;
yes,
acu
te t
rian
gle
;
1
0 2 +
11 2 <
20 2
(
5 √
"
2 ) 2
+ 1
0 2 >
11 2
n
o,
21.5
+ 2
4 <
55.5
y
es,
rig
ht
tria
ng
le;
yes,
rig
ht
tria
ng
le;
3
0 2 +
40 2 =
50 2
65 2 +
72 2 =
97 2
64
144
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 8-2
Ch
ap
ter
8
15
G
len
co
e G
eo
me
try
Wo
rd
Pro
ble
m P
racti
ce
Th
e P
yth
ag
ore
an
Th
eo
rem
an
d I
ts C
on
vers
e
1
. S
IDE
WA
LK
S C
on
stru
ctio
n w
ork
ers
are
bu
ild
ing a
marb
le s
idew
alk
aro
un
d a
p
ark
th
at
is s
hap
ed
lik
e a
rig
ht
tria
ngle
. E
ach
marb
le s
lab a
dd
s 2 f
eet
to t
he
len
gth
of
the s
idew
alk
. T
he w
ork
ers
fi
nd
th
at
exact
ly 1
071 a
nd
1840 s
labs
are
requ
ired
to m
ak
e t
he s
idew
alk
s alo
ng t
he s
hort
sid
es
of
the p
ark
. H
ow
m
an
y s
labs
are
requ
ired
to m
ak
e t
he
sid
ew
alk
th
at
run
s alo
ng t
he l
on
g s
ide
of
the p
ark
?
2
. R
IGH
T A
NG
LE
S C
lyd
e m
ak
es
a t
rian
gle
u
sin
g t
hre
e s
tick
s of
len
gth
s 20 i
nch
es,
21 i
nch
es,
an
d 2
8 i
nch
es.
Is
the t
rian
gle
a r
igh
t tr
ian
gle
? E
xp
lain
.
3
. T
ET
HE
RS
T
o h
elp
su
pp
ort
a f
lag p
ole
, a
50-f
oot-
lon
g t
eth
er
is t
ied
to t
he p
ole
at
a p
oin
t 40 f
eet
above t
he g
rou
nd
. T
he
teth
er
is p
ull
ed
tau
t an
d t
ied
to a
n
an
chor
in t
he g
rou
nd
. H
ow
far
aw
ay
from
th
e b
ase
of
the p
ole
is
the a
nch
or?
4
. FLIG
HT
A
n a
irp
lan
e l
an
ds
at
an
air
port
60 m
iles
east
an
d 2
5 m
iles
nort
h o
f w
here
it
took
off
.
60 m
i
25
mi
H
ow
far
ap
art
are
th
e t
wo a
irp
ort
s?
5
. P
YT
HA
GO
RE
AN
TR
IPLE
S M
s. J
on
es
ass
ign
ed
her
fift
h-p
eri
od
geom
etr
y c
lass
th
e f
oll
ow
ing p
roble
m.
Let
m a
nd
n b
e t
wo p
osi
tive i
nte
gers
w
ith
m >
n.
Let
a =
m2 –
n2,
b =
2m
n,
an
d c
= m
2 +
n2.
a.
Sh
ow
th
at
there
is
a r
igh
t tr
ian
gle
w
ith
sid
e l
en
gth
s a
, b,
an
d c
.
b.
Com
ple
te t
he f
oll
ow
ing t
able
.
mn
ab
c
21
34
5
31
86
10
32
512
13
41
15
817
42
12
16
20
43
724
25
51
24
10
26
c.
Fin
d a
Pyth
agore
an
tri
ple
th
at
corr
esp
on
ds
to a
rig
ht
tria
ngle
wit
h
a h
yp
ote
nu
se 2
52 =
625 u
nit
s lo
ng.
(Hin
t: U
se t
he t
able
you
com
ple
ted
fo
r E
xerc
ise b
to f
ind
tw
o p
osi
tive
inte
gers
m a
nd
n w
ith
m >
n a
nd
m
2 +
n2 =
625.)
8-2
6
5 m
i
2
129
3
0 f
t
n
o;
20 2 +
21 2 ≠
28 2
S
am
ple
an
sw
er:
a 2 +
b 2 =
( m
2 –
n 2 ) 2
+ (
2m
n) 2
= m
4 -
2 m
2 n
2 +
n 4 +
4 m
2 n
2 =
m 4 +
2 m
2 n
2 +
n 4
an
d c
2 =
( m
2 +
n 2 ) 2
= m
4 +
2 m
2 n
2 +
n 4 .
Th
is m
ean
s t
hat
a 2 +
b 2 =
c 2 ,
so
th
at
a,
b,
an
d c
do
fo
rm t
he s
ides o
f a r
igh
t
tria
ng
le b
y t
he c
on
vers
e o
f th
e
Pyth
ag
ore
an
Th
eo
rem
.
S
am
ple
an
sw
er:
Take m
= 2
4
an
d n
= 7
to
get
a =
527,
b =
336,
an
d c
= 6
25.
Answers (Lesson 8-2)
Answers
Co
pyri
gh
t ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f T
he
Mc
Gra
w-H
ill C
om
pa
nie
s,
Inc
.
Chapter 8 A7 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
16
G
len
co
e G
eo
me
try
En
rich
men
t
Co
nvers
e o
f a R
igh
t Tri
an
gle
Th
eo
rem
You
have l
earn
ed
th
at
the m
easu
re o
f th
e a
ltit
ud
e f
rom
th
e v
ert
ex o
f th
e r
igh
t an
gle
of
a r
igh
t tr
ian
gle
to i
ts h
yp
ote
nu
se i
s th
e g
eom
etr
ic
mean
betw
een
th
e m
easu
res
of
the t
wo s
egm
en
ts o
f th
e h
yp
ote
nu
se.
Is t
he c
on
vers
e o
f th
is t
heore
m t
rue?
In o
rder
to f
ind
ou
t, i
t w
ill
help
to
rew
rite
th
e o
rigin
al
theore
m i
n i
f-th
en
form
as
foll
ow
s.
If △
AB
Q i
s a r
igh
t tr
ian
gle
wit
h r
igh
t an
gle
at
Q,
then
QP
is
the g
eom
etr
ic m
ean
betw
een
AP
an
d P
B,
wh
ere
P
is b
etw
een
A a
nd
B a
nd
−−
−
QP
is
perp
en
dic
ula
r to
−−
AB
.
1
. W
rite
th
e c
on
vers
e o
f th
e i
f-th
en
form
of
the t
heore
m.
2
. Is
th
e c
on
vers
e o
f th
e o
rigin
al
theore
m t
rue?
Refe
r to
th
e f
igu
re a
t th
e r
igh
t to
exp
lain
you
r an
swer.
Y
ou
may f
ind
it
inte
rest
ing t
o e
xam
ine t
he o
ther
theore
ms
in
Ch
ap
ter
8 t
o s
ee w
heth
er
their
con
vers
es
are
tru
e o
r fa
lse.
You
wil
l n
eed
to r
est
ate
th
e t
heore
ms
care
full
y i
n o
rder
to w
rite
th
eir
co
nvers
es.
Q
BP
A
Q
BP
A
8-2
If
QP
is t
he g
eo
metr
ic m
ean
betw
een
AP
an
d
PB
, w
here
P i
s b
etw
een
A a
nd
B a
nd
−−
QP
⊥ −
−
AB
,
then
△A
BQ
is a
rig
ht
tria
ng
le w
ith
rig
ht
an
gle
at
Q.
Y
es;
(PQ
) 2 =
(A
P)(
PB
) im
plies t
hat
PQ
−
AP
= P
B
−
PQ
.
S
ince b
oth
∠A
PQ
an
d ∠
QP
B a
re r
igh
t an
gle
s,
they a
re c
on
gru
en
t. T
here
fore
△A
PQ
∼ △
QP
B b
y S
AS
sim
ilari
ty.
So
∠A
& ∠
PQ
B a
nd
∠A
QP
& ∠
B.
Bu
t th
e a
cu
te
an
gle
s o
f △
AQ
P a
re c
om
ple
men
tary
an
d
m∠
AQ
B =
m∠
AQ
P +
m∠
PQ
B.
Hen
ce
m∠
AQ
B =
90 a
nd
△A
QB
is a
rig
ht
tria
ng
le
wit
h r
igh
t an
gle
at
Q.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 8-2
Ch
ap
ter
8
17
G
len
co
e G
eo
me
try
Sp
read
sheet
Act
ivit
y
Pyth
ag
ore
an
Tri
ple
s
You
can
use
a s
pre
ad
sheet
to d
ete
rmin
e w
heth
er
thre
e w
hole
nu
mbers
fo
rm a
Pyth
agore
an
tri
ple
.
U
se
a s
pre
ad
sh
ee
t to
de
term
ine
wh
eth
er t
he
nu
mb
ers 1
2,
16
, a
nd
20
form
a P
yth
ag
ore
an
trip
le.
Ste
p 1
In
cell
A1,
en
ter
12.
In c
ell
B1,
en
ter
16 a
nd
in
cell
C1,
en
ter
20.
Th
e lo
nges
t si
de
shou
ld b
e en
tere
d i
n c
olu
mn
C.
Ste
p 2
In
cell
D1,
en
ter
an
equ
als
sig
n f
oll
ow
ed
by
IF(A
1^
2+
B1^
2=
C1^
2,“
YE
S”,
“NO
”).
Th
is w
ill
retu
rn “
YE
S”
if t
he s
et
of
nu
mbers
is
a
Pyth
agore
an
tri
ple
an
d w
ill
retu
rn “
NO
” if
it
is n
ot.
Th
e n
um
bers
12,
16,
an
d 2
0 f
orm
a P
yth
agore
an
tri
ple
.
U
se
a s
pre
ad
sh
ee
t to
de
term
ine
wh
eth
er t
he
nu
mb
ers 3
, 6
, a
nd
12
form
a P
yth
ag
ore
an
trip
le.
Ste
p 1
In
cell
A2,
en
ter
3,
in c
ell
B2,
en
ter
6,
an
d i
n c
ell
C2,
en
ter
12.
Ste
p 2
C
lick
on
th
e b
ott
om
rig
ht
corn
er
of
cell
D1 a
nd
dra
g i
t to
D2.
Th
is
wil
l d
ete
rmin
e w
heth
er
or
not
the s
et
of
nu
mbers
is
a P
yth
agore
an
tr
iple
.
Th
e n
um
bers
3,
6,
an
d 1
2 d
o n
ot
form
a P
yth
agore
an
tri
ple
.
Exerc
ises
Use
a s
pre
ad
sh
ee
t to
de
term
ine
wh
eth
er e
ach
se
t o
f n
um
be
rs
form
s a
Py
tha
go
re
an
trip
le.
1. 14,
48,
50
2
. 16,
30,
34
3
. 5,
5,
9
4
. 4,
5,
7
5. 18,
24,
30
6. 10,
24,
26
7
. 25,
60,
65
8
. 2,
4,
5
9. 19,
21,
22
10
. 18,
80,
82
1
1. 5,
12,
13
1
2. 20,
48,
52
8-2
Exam
ple
1
Exam
ple
2
y
es
yes
no
n
o
yes
yes
y
es
no
n
o
y
es
yes
yes
Answers (Lesson 8-2)
Co
pyrig
ht ©
Gle
nc
oe
/Mc
Gra
w-H
ill, a d
ivis
ion
of T
he
Mc
Gra
w-H
ill Co
mp
an
ies, In
c.
Chapter 8 A8 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
18
G
len
co
e G
eo
me
try
Stu
dy
Gu
ide a
nd
In
terv
en
tio
n
Sp
ecia
l R
igh
t Tri
an
gle
s
Pro
pe
rtie
s o
f 4
5°-
45
°-9
0°
Tri
an
gle
s T
he s
ides
of
a 4
5°-
45°-
90°
righ
t tr
ian
gle
have a
sp
eci
al
rela
tion
ship
.
If
th
e l
eg
of
a 4
5°-
45
°-9
0°
rig
ht
tria
ng
le i
s x
un
its,
sh
ow
tha
t th
e h
yp
ote
nu
se
is x
√ "
2
un
its.
x√
x
x2
45
°
45
°
Usi
ng t
he P
yth
agore
an
Th
eore
m w
ith
a
= b
= x
, th
en
c2
= a
2 +
b2
c2
= x
2 +
x2
c2
=
2x
2
c
= √
##
2x
2
c
= x
√ #
2
In
a 4
5°-
45
°-9
0°
rig
ht
tria
ng
le t
he
hy
po
ten
use
is √
"
2 t
ime
s
the
le
g.
If t
he
hy
po
ten
use
is 6
un
its,
fin
d t
he
le
ng
th o
f e
ach
le
g.
Th
e h
yp
ote
nu
se i
s √
#
2 t
imes
the l
eg,
so
div
ide t
he l
en
gth
of
the h
yp
ote
nu
se b
y √
#
2 .
a =
6
−
√ #
2
=
6
−
√ #
2 .
√ #
2
−
√ #
2
=
6 √
#
2
−
2
=
3 √
#
2 u
nit
s
Exerc
ises
Fin
d x
.
1.
x
8
45
°
45
°
2.
x
45
°3
√2
3.
45°
4
x
4.
xx
18
5.
45°
16
x
x
6.
45°
24
x
√2
7. If
a 4
5°-
45°-
90°
tria
ngle
has
a h
yp
ote
nu
se l
en
gth
of
12,
fin
d t
he l
eg l
en
gth
.
8
. D
ete
rmin
e t
he l
en
gth
of
the l
eg o
f 45°-
45°-
90°
tria
ngle
wit
h a
hyp
ote
nu
se l
en
gth
of
25 i
nch
es.
9. F
ind
th
e l
en
gth
of
the h
yp
ote
nu
se o
f a 4
5°-
45°-
90°
tria
ngle
wit
h a
leg l
en
gth
of
14 c
en
tim
ete
rs.
8-3
Exam
ple
1Exam
ple
2
9 √
"
2 ≈
12.7
8 √
"
2 ≈
11.3
24
8 √
"
2 ≈
11.3
3
4 √
"
2 ≈
5.7
6 √
"
2 ≈
8.5
25
√ "
2
−
2
in
. ≈
17.7
in
.
14 √
"
2 c
m ≈
19.8
cm
Lesson 8-3
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
19
G
len
co
e G
eo
me
try
Stu
dy
Gu
ide a
nd
In
terv
en
tio
n
(con
tin
ued
)
Sp
ecia
l R
igh
t Tri
an
gle
s
Pro
pe
rtie
s o
f 3
0°-
60
°-9
0°
Tri
an
gle
s T
he s
ides
of
a 3
0°-
60°-
90°
righ
t tr
ian
gle
als
o
have a
sp
eci
al
rela
tion
ship
.
In
a 3
0°-
60
°-9
0°
rig
ht
tria
ng
le t
he
hy
po
ten
use
is t
wic
e t
he
sh
orte
r l
eg
. S
ho
w t
ha
t th
e l
on
ge
r l
eg
is √
"
3 t
ime
s
the
sh
orte
r l
eg
.
△ M
NQ
is
a 3
0°-
60°-
90°
righ
t tr
ian
gle
, an
d t
he l
en
gth
of
the
hyp
ote
nu
se −
−−
MN
is
two t
imes
the l
en
gth
of
the s
hort
er
sid
e −−
− N
Q .
Use
th
e P
yth
agore
an
Th
eore
m.
a2 =
(2
x)2
- x
2
a2 =
c2 -
b2
a2 =
4x
2 -
x2
M
ultip
ly.
a2 =
3x
2
Subtr
act.
a =
√ #
#
3x
2
Take t
he p
ositiv
e s
quare
root
of
each s
ide.
a =
x √
#
3
Sim
plif
y.
In
a 3
0°-
60
°-9
0°
rig
ht
tria
ng
le,
the
hy
po
ten
use
is 5
ce
nti
me
ters.
Fin
d t
he
le
ng
ths o
f th
e o
the
r t
wo
sid
es o
f th
e t
ria
ng
le.
If t
he h
yp
ote
nu
se o
f a 3
0°-
60°-
90°
righ
t tr
ian
gle
is
5 c
en
tim
ete
rs,
then
th
e l
en
gth
of
the
short
er
leg i
s on
e-h
alf
of
5,
or
2.5
cen
tim
ete
rs.
Th
e l
en
gth
of
the l
on
ger
leg i
s √
#
3 t
imes
the
len
gth
of
the s
hort
er
leg,
or
(2.5
)( √
#
3 )
cen
tim
ete
rs.
Exerc
ises
Fin
d x
an
d y
.
1.
x y30
°
60
°1 2
2.
x
y
60
°
8
3.
x
y11
30
°
4.
xy
30
°
9√
3
5.
xy
60
°
12
6.
60°
x20
y
7. A
n e
qu
ilate
ral
tria
ngle
has
an
alt
itu
de l
en
gth
of
36 f
eet.
Dete
rmin
e t
he l
en
gth
of
a s
ide
of
the t
rian
gle
.
8
. F
ind
th
e l
en
gth
of
the s
ide o
f an
equ
ilate
ral
tria
ngle
th
at
has
an
alt
itu
de l
en
gth
of
45 c
en
tim
ete
rs.
8-3
30°
60°
x2
xaExam
ple
1
Exam
ple
2
x =
1;
x =
8 √
"
3 ≈
13.9
;
x =
5.5
;
y =
0.5
√ "
3 ≈
0.9
y =
16
y =
5.5
√ "
3 ≈
9.5
x =
9;
x =
4 √
"
3 ≈
6.9
;
x =
10 √
"
3 ≈
17.3
;
y =
18
y =
8 √
"
3 ≈
13.9
y
= 1
0
24 √
"
3 f
eet
≈ 4
1.6
ft
30 √
"
3 c
m ≈
52 c
m
Answers (Lesson 8-3)
Answers
Co
pyri
gh
t ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f T
he
Mc
Gra
w-H
ill C
om
pa
nie
s,
Inc
.
Chapter 8 A9 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
20
G
len
co
e G
eo
me
try
8-3
Sk
ills
Practi
ce
Sp
ecia
l R
igh
t Tri
an
gle
s
Fin
d x
.
1.
45°
25
x
2
.
45°
17
x
3
. 45°
48
x
4
.
45°
100
x
5
.
45°
100
x
6
.
45°88
x
7. D
ete
rmin
e t
he l
en
gth
of
the l
eg o
f 45°-
45°-
90°
tria
ngle
wit
h a
hypote
nu
se l
en
gth
of
26.
8. F
ind
th
e l
en
gth
of
the h
yp
ote
nu
se o
f a 4
5°-
45°-
90°
tria
ngle
wit
h a
leg l
en
gth
of
50 c
en
tim
ete
rs.
Fin
d x
an
d y
.
9.
30°
x11
y
1
0.
60°
x
8
y
√3
1
1.
30°
x
5
y
√3
12
.
60°
x
30
y
1
3.
30°
x
21
y√3
1
4.
60°
x
52
y√3
15. A
n e
qu
ilate
ral
tria
ngle
has
an
alt
itu
de l
en
gth
of
27 f
eet.
Dete
rmin
e t
he l
en
gth
of
a s
ide
of
the t
rian
gle
.
16
. F
ind
th
e l
en
gth
of
the s
ide o
f an
equ
ilate
ral
tria
ngle
th
at
has
an
alt
itu
de l
en
gth
of
11 √
"
3 f
eet.
2
5 √
"
2
8.5
√ "
2 o
r 1
7 √
"
2
−
2
4
8 √
"
2
2
2;
11 √
"
3
12;
4 √
"
3
15;
10 √
"
3
5
0 √
"
2
100 √
"
2
44 √
"
2
13 √
"
2
5
0 √
"
2 c
m
1
5;
15 √
"
3
42;
21
78;
26 √
"
3
1
8 √
"
3
2
2
Lesson 8-3
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
21
G
len
co
e G
eo
me
try
Fin
d x
.
1.
45°
14
x
2.
45°
45
x
3.
45°
22
x
4.
45°
210
x
5.
45°
88
x
6.
45°
x5
√2
Fin
d x
an
d y
.
7.
30°
x9
y
8.
60°
x
4
y
√3
9.
30°
x
20
y
1
0.
60°
x
98
y
11
. D
ete
rmin
e t
he l
en
gth
of
the l
eg o
f 45°-
45°-
90°
tria
ngle
wit
h a
hyp
ote
nu
se l
en
gth
of
38.
12. F
ind
th
e l
en
gth
of
the h
yp
ote
nu
se o
f a 4
5°-
45°-
90°
tria
ngle
wit
h a
leg l
en
gth
of
77 c
en
tim
ete
rs.
13. A
n e
qu
ilate
ral
tria
ngle
has
an
alt
itu
de l
en
gth
of
33 f
eet.
Dete
rmin
e t
he l
en
gth
of
a s
ide
of
the t
rian
gle
.
14
. B
OT
AN
ICA
L G
AR
DE
NS
O
ne o
f th
e d
isp
lays
at
a b
ota
nic
al
gard
en
is a
n h
erb
gard
en
pla
nte
d i
n t
he s
hap
e o
f a s
qu
are
. T
he s
qu
are
measu
res
6 y
ard
s on
each
sid
e.
Vis
itors
can
vie
w t
he h
erb
s fr
om
a
dia
gon
al
path
way t
hro
ugh
th
e g
ard
en
. H
ow
lon
g i
s th
e p
ath
way?
6 yd
6 yd
6 yd
6 yd
Practi
ce
Sp
ecia
l R
igh
t Tri
an
gle
s
8-3
1
4 √
"
2
22.5
√ "
2 o
r 4
5 √
"
2
−
2
2
2 √
"
2
1
05 √
"
2
88 √
"
2
10
x
= 1
8;
x
= 6
;
y
= 9
√ "
3
y =
2 √
"
3
x
= 2
0 √
"
3 ;
y
= 4
0
1
9 √
"
2
7
7 √
"
2 c
m
2
2 √
"
3 f
t
6
√ "
2 y
d o
r ab
ou
t 8.4
9 y
d
x =
49;
y =
49 √
"
3
Answers (Lesson 8-3)
Co
pyrig
ht ©
Gle
nc
oe
/Mc
Gra
w-H
ill, a d
ivis
ion
of T
he
Mc
Gra
w-H
ill Co
mp
an
ies, In
c.
Chapter 8 A10 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
22
G
len
co
e G
eo
me
try
Wo
rd
Pro
ble
m P
racti
ce
Sp
ecia
l R
igh
t Tri
an
gle
s
1
. O
RIG
AM
I A
squ
are
pie
ce o
f p
ap
er
150 m
illi
mete
rs o
n a
sid
e i
s fo
lded
in
half
alo
ng a
dia
gon
al.
Th
e r
esu
lt i
s a
45°-
45°-
90°
tria
ngle
. W
hat
is t
he l
en
gth
of
the h
yp
ote
nu
se o
f th
is t
rian
gle
?
2
. E
SC
ALA
TO
RS
A
40-f
oot-
lon
g e
scala
tor
rise
s fr
om
th
e f
irst
flo
or
to t
he s
eco
nd
floor
of
a s
hop
pin
g m
all
. T
he e
scala
tor
mak
es
a 3
0°
an
gle
wit
h t
he h
ori
zon
tal.
40 f
t
30
°
? f
t
H
ow
hig
h a
bove t
he f
irst
flo
or
is t
he
seco
nd
flo
or?
3
. H
EX
AG
ON
S A
box o
f ch
oco
late
s sh
ap
ed
lik
e a
regu
lar
hexagon
is
pla
ced
sn
ugly
insi
de o
f a r
ect
an
gu
lar
box a
s sh
ow
n i
n
the f
igu
re.
If
th
e s
ide l
en
gth
of
the h
exagon
is
3 i
nch
es,
wh
at
are
th
e d
imen
sion
s of
the r
ect
an
gu
lar
box?
4
. W
IND
OW
S A
larg
e s
tain
ed
gla
ss
win
dow
is
con
stru
cted
fro
m s
ix 3
0°-
60°-
90°
tria
ngle
s as
show
n i
n t
he f
igu
re.
3 m
W
hat
is t
he h
eig
ht
of
the w
ind
ow
?
5
. M
OV
IES
K
im a
nd
Yola
nd
a a
re
watc
hin
g a
movie
in
a m
ovie
th
eate
r.
Yola
nd
a i
s si
ttin
g x
feet
from
th
e s
creen
an
d K
im i
s 15 f
eet
beh
ind
Yola
nd
a.
30
°45
°
15 f
tx
ft
T
he a
ngle
th
at
Kim
’s l
ine o
f si
gh
t
to t
he t
op
of
the s
creen
mak
es
wit
h
the h
ori
zon
tal
is 3
0°.
Th
e a
ngle
th
at
Yola
nd
a’s
lin
e o
f si
gh
t to
th
e t
op
of
the
scre
en
mak
es
wit
h t
he h
ori
zon
tal
is 4
5°.
a.
How
hig
h i
s th
e t
op
of
the s
creen
in
term
s of
x?
b.
Wh
at
is x
+ 1
5
−
x
?
c.
How
far
is Y
ola
nd
a f
rom
th
e s
creen
?
Rou
nd
you
r an
swer
to t
he n
eare
st
ten
th.
8-3
1
50 √
#
2 m
m
6
in
. b
y 3
√ #
3 i
n.
2
0ft
4 √
#
3
−3
m ≈
2.3
1 m
√ #
3 f
t
x f
t
20.5
ft
Lesson 8-3
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
23
G
len
co
e G
eo
me
try
En
ric
hm
en
t
√1
1
1
3√
2
Co
nstr
ucti
ng
Valu
es o
f S
qu
are
Ro
ots
Th
e d
iagra
m a
t th
e r
igh
t sh
ow
s a r
igh
t is
osc
ele
s tr
ian
gle
wit
h
two l
egs
of
len
gth
1 i
nch
. B
y t
he P
yth
agore
an
Th
eore
m,
the l
en
gth
of
the h
yp
ote
nu
se i
s √
$
2 i
nch
es.
By c
on
stru
ctin
g a
n a
dja
cen
t ri
gh
t
tria
ngle
wit
h l
egs
of
√ $
2
inch
es
an
d 1
in
ch,
you
can
cre
ate
a s
egm
en
t
of
len
gth
√ $
3 .
By c
on
tin
uin
g t
his
pro
cess
as
show
n b
elo
w,
you
can
con
stru
ct a
“wh
eel”
of
squ
are
roots
. T
his
wh
eel
is c
all
ed
th
e “
Wh
eel
of
Th
eod
oru
s”
aft
er
a G
reek
ph
iloso
ph
er
wh
o l
ived
abou
t 400 B
.C.
Con
tin
ue c
on
stru
ctin
g t
he w
heel
un
til
you
mak
e a
segm
en
t of
len
gth
√ $
$
18 .
1
1
11
1
1√ 2
√ 3
√5
√ 6
√ 7 √ 8
10
√$
√$
11
√$ 12
√$ 13
√$ 14
√$ 15
√$ 17
√$ 18
√$ 16
= 4
√ 4
= 2
√ 9
= 3
8-3
Answers (Lesson 8-3)
Answers
Co
pyri
gh
t ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f T
he
Mc
Gra
w-H
ill C
om
pa
nie
s,
Inc
.
Chapter 8 A11 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
24
G
len
co
e G
eo
me
try
Tri
go
no
me
tric
Ra
tio
s T
he r
ati
o o
f th
e l
en
gth
s of
two s
ides
of
a r
igh
t tr
ian
gle
is
call
ed
a t
rig
on
om
etr
ic r
ati
o.
Th
e t
hre
e m
ost
com
mon
rati
os
are
sin
e,
co
sin
e,
an
d t
an
ge
nt,
wh
ich
are
abbre
via
ted
sin
, co
s, a
nd
ta
n,
resp
ect
ively
.
sin
R =
le
g o
pp
osi
te ∠
R
−
hyp
ote
nu
se
co
s R
=
leg a
dja
cen
t to
∠R
−
−
h
yp
ote
nu
se
ta
n R
=
leg o
pp
osi
te ∠
R
−
−
le
g a
dja
cen
t to
∠R
=
r −
t =
s −
t =
r −
s
F
ind
sin
A,
co
s A
, a
nd
ta
n A
. E
xp
re
ss e
ach
ra
tio
as
a f
ra
cti
on
an
d a
de
cim
al
to t
he
ne
are
st
hu
nd
re
dth
.
sin
A =
op
posi
te l
eg
−
hyp
ote
nu
se
cos
A =
ad
jace
nt
leg
−
h
yp
ote
nu
se
tan
A =
op
posi
te l
eg
−
ad
jace
nt
leg
=
BC
−
BA
=
AC
−
AB
=
BC
−
AC
=
5
−
13
= 1
2
−
13
=
5
−
12
≈ 0
.38
≈ 0
.92
≈ 0
.42
Exerc
ises
Fin
d s
in J
, co
s J
, ta
n J
, sin
L,
co
s L
, a
nd
ta
n L
. E
xp
re
ss e
ach
ra
tio
as a
fra
cti
on
an
d a
s a
de
cim
al
to t
he
ne
are
st
hu
nd
re
dth
if
ne
ce
ssa
ry
.
1.
2.
3.
1213
5 CB
A
Stu
dy
Gu
ide a
nd
In
terv
en
tio
n
Trigonometry
8-4
s
tr TS
R
Exam
ple
20
12
16
40
24
32
36
12
√3
24
√3
s
in J
= 1
2
−
20 =
0.6
;
c
os
J =
16
−
20 =
0.8
;
ta
n J
= 1
2
−
16 =
0.7
5;
s
in L
= 1
6
−
20 =
0.8
;
c
os
L=
12
− 20
= 0
.6;
ta
n L
= 1
6
−
12 ≈
1.3
3
s
in J
= 2
4
−
40 =
0.6
;
c
os
J =
32
−
40 =
0.8
;
ta
n J
= 2
4
−
32 =
0.7
5;
s
in L
= 3
2
−
40 =
0.8
;
c
os
L =
24
−
32 =
0.6
;
ta
n L
= 3
2
−
24 ≈
1.3
3
s
in J
=
36
−
24 √
$
3
≈ 0
.87
;
c
os
J =
12 √
$
3
−
24 √
$
3
= 0
.5;
ta
n J
=
36
−
12 √
$
3
≈ 1
.73
;
s
in L
= 1
2 √
$
3
−
24 √
$
3
= 0
.5;
c
os
L =
36
−
24 √
$
3
≈ 0
.87
;
ta
n L
= 1
2 √
$
3
−
36
≈ 0
.58
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 8-4
Ch
ap
ter
8
25
G
len
co
e G
eo
me
try
Stu
dy
Gu
ide a
nd
In
terv
en
tio
n
(con
tin
ued
)
Trigonometry
Use
In
ve
rse
Tri
go
no
me
tric
Ra
tio
s Y
ou
can
use
a c
alc
ula
tor
an
d t
he s
ine,
cosi
ne,
or
tan
gen
t to
fin
d t
he m
easu
re o
f th
e a
ngle
, ca
lled
th
e i
nv
erse
of
the t
rigon
om
etr
ic r
ati
o.
U
se
a c
alc
ula
tor t
o f
ind
th
e m
ea
su
re
of
∠T
to
th
e n
ea
re
st
ten
th.
Th
e m
easu
res
giv
en
are
th
ose
of
the l
eg o
pp
osi
te ∠
T a
nd
th
e
hyp
ote
nu
se,
so w
rite
an
equ
ati
on
usi
ng t
he s
ine r
ati
o.
sin
T =
op
p
−
hyp
sin
T =
2
9
−
34
If s
in T
= 2
9
−
34 ,
then
sin
-1 2
9
−
34 =
m∠
T.
Use
a c
alc
ula
tor.
So,
m∠
T ≈
58.5
.
Exerc
ises
Use
a c
alc
ula
tor t
o f
ind
th
e m
ea
su
re
of
∠T
to
th
e n
ea
re
st
ten
th.
1.
34
14
√3
2.
7
18
3.
87
34
4.
101
32
5.
67
10
√3
6.
39
14
√2
8-4 Exam
ple
34
29
71.5
°75°
30.5
°
35.5
°22.9
°67°
Answers (Lesson 8-4)
Co
pyrig
ht ©
Gle
nc
oe
/Mc
Gra
w-H
ill, a d
ivis
ion
of T
he
Mc
Gra
w-H
ill Co
mp
an
ies, In
c.
Chapter 8 A12 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
26
G
len
co
e G
eo
me
try
Fin
d s
in R
, co
s R
, ta
n R
, sin
S,
co
s S
, a
nd
ta
n S
.
Ex
pre
ss e
ach
ra
tio
as a
fra
cti
on
an
d a
s a
de
cim
al
to t
he
ne
are
st
hu
nd
re
dth
.
1. r =
16, s =
30, t
= 3
4
2. r =
10, s =
24, t
= 2
6
Use
a s
pe
cia
l rig
ht
tria
ng
le t
o e
xp
re
ss e
ach
trig
on
om
etr
ic r
ati
o a
s a
fra
cti
on
an
d
as a
de
cim
al
to t
he
ne
are
st
hu
nd
re
dth
if
ne
ce
ssa
ry
.
3. si
n 3
0°
4. ta
n 4
5°
5. co
s 60°
6. si
n 6
0°
7. ta
n 3
0°
8. co
s 45°
Fin
d x
. R
ou
nd
to
th
e n
ea
re
st
hu
nd
re
dth
if
ne
ce
ssa
ry
.
9.
7x
36°
10.
11.
Use
a c
alc
ula
tor t
o f
ind
th
e m
ea
su
re
of ∠B
to
th
e n
ea
re
st
ten
th.
12.
13.
14.
sR
S Trt
Sk
ills
Practi
ce
Trigonometry
8-4
12
x
15°
6
x°
18
15
x°
12
22
x°
19
12
x
63°
sin
R =
8
−
17 ≈
0.4
7;
sin
R =
5
−
13 ≈
0.3
8;
co
s R
= 1
5
−
17 ≈
0.8
8;
co
s R
= 1
2
−
13 ≈
0.9
2;
tan
R =
8
−
15 ≈
0.5
3;
ta
n R
=
5
−
12 ≈
0.4
2;
sin
S =
15
−
17 ≈
0.8
8;
sin
S =
12
−
13 ≈
0.9
2;
co
s S
=
8
−
17 ≈
0.4
7;
co
s S
=
5
−
13 ≈
0.3
8;
tan
S =
15
−
8
≈ 1
.88
tan
S =
12
−
5
≈ 2
.4
x =
5.0
9x =
13.4
7x =
3.1
1
49.2
°70.5
°36.9
°
1
−
2 ;
0.5
1 1
−
2 ;
0.5
√ $
3
−
2
; 0.8
7
√ $
3
−
3
; 0.5
8
√ $
2
−
2
; 0.7
1
Lesson 8-4
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
27
G
len
co
e G
eo
me
try
Fin
d s
in L
, co
s L
, ta
n L
, sin
M,
co
s M
, a
nd
ta
n M
.
Ex
pre
ss e
ach
ra
tio
as a
fra
cti
on
an
d a
s a
de
cim
al
to t
he
ne
are
st
hu
nd
re
dth
.
1.
ℓ =
15, m
= 3
6, n
= 3
9
2.
ℓ =
12, m
= 1
2 √
#
3 , n
= 2
4
Fin
d x
. R
ou
nd
to
th
e n
ea
re
st
hu
nd
re
dth
.
3
.
11
64
°
x
4
.
5.
41
°x
32
Use
a c
alc
ula
tor t
o f
ind
th
e m
ea
su
re
of
∠B
to
th
e n
ea
re
st
ten
th.
6
.
8
14
7
.
30
25
8
.
9
. G
EO
GR
AP
HY
D
iego u
sed
a t
heod
oli
te t
o m
ap
a r
egio
n o
f la
nd
for
his
class
in
geom
orp
holo
gy.
To d
ete
rmin
e t
he e
levati
on
of
a v
ert
ical
rock
form
ati
on
, h
e m
easu
red
th
e d
ista
nce
fro
m t
he b
ase
of
the f
orm
ati
on
to
his
posi
tion
an
d t
he a
ngle
betw
een
th
e g
rou
nd
an
d t
he l
ine o
f si
gh
t to
th
e
top
of
the f
orm
ati
on
. T
he d
ista
nce
was
43 m
ete
rs a
nd
th
e a
ngle
was
36°.
Wh
at
is t
he h
eig
ht
of
the f
orm
ati
on
to t
he n
eare
st m
ete
r?
29
29
°x
36
° 43m
ML
N
m
n
Practi
ce
Trigonometry
8-4
739
sin
L =
5
−
13 ≈
0.3
8;
sin
L =
1
−
2 =
0.5
0;
co
s L
= 1
2
−
13 ≈
0.9
2;
co
s L
= √
$
3
−
2 ≈
0.8
7;
tan
L =
5
−
12 ≈
0.4
2;
ta
n L
=
1
−
√ $
3
or
√ $
3
−
3 ≈
0.5
8;
sin
M =
12
−
13 ≈
0.9
2;
sin
M =
√ $
3
−
2 ≈
0.8
7;
co
s M
=
5
−
13 ≈
0.3
8;
co
s M
= 1
−
2 =
0.5
0;
tan
M =
12
−
5
≈ 2
.4
tan
M =
√ $
3 ≈
1.7
3
60.3
°33.6
°79.7
°
22.5
525.3
624.1
5
31 m
Answers (Lesson 8-4)
Answers
Co
pyri
gh
t ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f T
he
Mc
Gra
w-H
ill C
om
pa
nie
s,
Inc
.
Chapter 8 A13 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
28
G
len
co
e G
eo
me
try
Wo
rd
Pro
ble
m P
racti
ce
Tri
go
no
metr
y
1
. R
AD
IO T
OW
ER
S K
ay i
s st
an
din
g n
ear
a 2
00-f
oot-
hig
h r
ad
io t
ow
er.
200 f
t ?
ft
49
°
U
se t
he i
nfo
rmati
on
in
th
e f
igu
re t
o
dete
rmin
e h
ow
far
Kay i
s fr
om
th
e t
op
of
the t
ow
er.
Exp
ress
you
r an
swer
as
a
trig
on
om
etr
ic f
un
ctio
n.
2
. R
AM
PS
A
60-f
oot
ram
p r
ises
from
th
e
firs
t fl
oor
to t
he s
eco
nd
flo
or
of
a p
ark
ing
gara
ge.
Th
e r
am
p m
ak
es
a 1
5°
an
gle
w
ith
th
e g
rou
nd
.
? f
t
60 f
t
15
°
H
ow
hig
h a
bove t
he f
irst
flo
or
is t
he
seco
nd
flo
or?
Exp
ress
you
r an
swer
as
a
trig
on
om
etr
ic f
un
ctio
n.
3
. T
RIG
ON
OM
ET
RY
M
eli
nd
a a
nd
Walt
er
were
both
solv
ing t
he s
am
e t
rigon
om
etr
y
pro
ble
m.
How
ever,
aft
er
they f
inis
hed
th
eir
com
pu
tati
on
s, M
eli
nd
a s
aid
th
e
an
swer
was
52 s
in 2
7°
an
d W
alt
er
said
th
e a
nsw
er
was
52 c
os
63
°. C
ou
ld t
hey
both
be c
orr
ect
? E
xp
lain
.
4
. LIN
ES
Jasm
ine d
raw
s li
ne m
on
a
coord
inate
pla
ne.
m x
y
O-5
5
5
W
hat
an
gle
does
m m
ak
e w
ith
th
e
x-a
xis
? R
ou
nd
you
r an
swer
to t
he
neare
st d
egre
e.
5
. N
EIG
HB
OR
S A
my,
Barr
y,
an
d C
hri
s li
ve
on
th
e s
am
e b
lock
. C
hri
s li
ves
up
th
e
stre
et
an
d a
rou
nd
th
e c
orn
er
from
Am
y,
an
d B
arr
y l
ives
at
the c
orn
er
betw
een
A
my a
nd
Ch
ris.
Th
e t
hre
e h
om
es
are
th
e
vert
ices
of
a r
igh
t tr
ian
gle
.
64
°
Bar
ryChri
s
Am
y
26
°
a.
Giv
e t
wo t
rigon
om
etr
ic e
xp
ress
ion
s fo
r th
e r
ati
o o
f B
arr
y’s
dis
tan
ce f
rom
A
my t
o C
hri
s’ d
ista
nce
fro
m A
my.
b.
Giv
e t
wo t
rigon
om
etr
ic e
xp
ress
ion
s fo
r th
e r
ati
o o
f B
arr
y’s
dis
tan
ce f
rom
C
hri
s to
Am
y’s
dis
tan
ce f
rom
Ch
ris.
c.
Giv
e a
tri
gon
om
etr
ic e
xp
ress
ion
for
the r
ati
o o
f A
my’s
dis
tan
ce f
rom
B
arr
y t
o C
hri
s’ d
ista
nce
fro
m B
arr
y.
8-4
20
0
−
sin
49
°
60 s
in 1
5°
Yes,
they a
re b
oth
co
rrect.
B
ecau
se 2
7 +
63
= 9
0,
the
sin
e o
f 27°
is t
he s
am
e r
ati
o a
s
the c
osin
e o
f 63°.
co
s 2
6°
or
sin
64°
co
s 6
4°
or
sin
26°
tan
64°
18°
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Lesson 8-4
Ch
ap
ter
8
29
G
len
co
e G
eo
me
try
En
ric
hm
en
t
Sin
e a
nd
Co
sin
e o
f A
ng
les
Th
e f
oll
ow
ing d
iagra
m c
an
be u
sed
to o
bta
in a
pp
roxim
ate
valu
es
for
the s
ine a
nd
cosi
ne o
f an
gle
s fr
om
0°
to 9
0°.
Th
e r
ad
ius
of
the c
ircl
e i
s 1.
So,
the s
ine a
nd
cosi
ne v
alu
es
can
be
read
dir
ect
ly f
rom
th
e v
ert
ical
an
d h
ori
zon
tal
axes.
F
ind
ap
pro
xim
ate
va
lue
s f
or s
in 4
0°
an
d
co
s 4
0°.
Co
nsid
er t
he
tria
ng
le f
orm
ed
by
th
e s
eg
me
nt
ma
rk
ed
40
°, a
s i
llu
str
ate
d b
y t
he
sh
ad
ed
tria
ng
le a
t rig
ht.
sin
40°
= a
−
c ≈
0.6
4
−
1
or
0.6
4
cos
40°
= b
−
c ≈
0.7
7
−
1
or
0.7
7
1
. U
se t
he d
iagra
m a
bove t
o c
om
ple
te t
he c
hart
of
valu
es.
x°
0°
10
°20
°30°
40
°50
°60
°70
°80
°90
°
sin x
°0
0.1
70.3
40.5
0.6
40.7
70.8
70.9
40.9
81
cos x
°1
0.9
80.9
40.8
70.7
70.6
40.5
0.3
40.1
70
2
. C
om
pare
th
e s
ine a
nd
cosi
ne o
f tw
o c
om
ple
men
tary
an
gle
s (a
ngle
s w
ith
a
sum
of
90°)
. W
hat
do y
ou
noti
ce?
00.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.91
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
90
°
0°
10
°
20
°
30
°
40
°
50
°
60
°
70
°
80
°
1 0
40
°0
.64
c=
1 u
nit
x°
b=
co
s x°
0.7
71
a=
sin
x°
Exam
ple
8-4 T
he s
ine o
f an
an
gle
is e
qu
al
to t
he c
osin
e o
f th
e
co
mp
lem
en
t o
f th
e a
ng
le.
Answers (Lesson 8-4)
Co
pyrig
ht ©
Gle
nc
oe
/Mc
Gra
w-H
ill, a d
ivis
ion
of T
he
Mc
Gra
w-H
ill Co
mp
an
ies, In
c.
Chapter 8 A14 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
30
G
len
co
e G
eo
me
try
Stu
dy
Gu
ide a
nd
In
terv
en
tio
n
An
gle
s o
f E
levati
on
an
d D
ep
ressio
n
An
gle
s o
f E
lev
ati
on
an
d D
ep
ress
ion
M
an
y r
eal-
worl
d p
roble
ms
that
involv
e l
ook
ing u
p t
o a
n o
bje
ct c
an
be d
esc
ribed
in
term
s of
an
an
gle
o
f e
lev
ati
on
, w
hic
h i
s th
e a
ngle
betw
een
an
obse
rver’
s li
ne o
f si
gh
t an
d a
hori
zon
tal
lin
e.
Wh
en
an
obse
rver
is l
ook
ing d
ow
n,
the a
ng
le o
f d
ep
re
ssio
n i
s th
e
an
gle
betw
een
th
e o
bse
rver’
s li
ne o
f si
gh
t an
d a
hori
zon
tal
lin
e.
T
he
an
gle
of
ele
va
tio
n f
ro
m p
oin
t A
to
th
e t
op
of
a c
liff
is 3
4°.
If
po
int A
is 1
00
0 f
ee
t fr
om
th
e b
ase
of
the
cli
ff,
ho
w h
igh
is t
he
cli
ff?
Let
x =
th
e h
eig
ht
of
the c
liff
.
ta
n 3
4°
=
x
−
1000
tan =
opposite
−
adja
cent
1000(t
an
34°)
= x
M
ultip
ly e
ach s
ide b
y 1
000.
674.5
≈ x
U
se a
calc
ula
tor.
Th
e h
eig
ht
of
the c
liff
is
abou
t 674.5
feet.
Exerc
ises
1
. H
ILL T
OP
T
he a
ngle
of
ele
vati
on
fro
m p
oin
t A
to t
he t
op
of
a h
ill
is 4
9°.
If
poin
t A
is
400 f
eet
from
th
e b
ase
of
the h
ill,
how
hig
h i
s th
e h
ill?
2. S
UN
F
ind
th
e a
ngle
of
ele
vati
on
of
the S
un
wh
en
a 1
2.5
-mete
r-ta
ll
tele
ph
on
e p
ole
cast
s an
18-m
ete
r-lo
ng s
had
ow
.
3. S
KII
NG
A
sk
i ru
n i
s 1000 y
ard
s lo
ng w
ith
a v
ert
ical
dro
p o
f 208 y
ard
s. F
ind
th
e a
ngle
of
dep
ress
ion
fro
m t
he t
op
of
the s
ki
run
to t
he b
ott
om
.
4
. A
IR T
RA
FFIC
F
rom
th
e t
op
of
a 1
20-f
oot-
hig
h t
ow
er,
an
air
tra
ffic
con
troll
er
obse
rves
an
air
pla
ne o
n t
he r
un
way
at
an
an
gle
of
dep
ress
ion
of
19°.
How
far
from
th
e b
ase
of
the t
ow
er
is t
he a
irp
lan
e?
18 m
12.5
mSun
?
400 f
t
?
49°
A
?
1000 f
t
34°
x
angle
of
elev
atio
n
line
of s
ight
8-5 Exam
ple
Ylin
e of
sig
ht
hori
zonta
lan
gle
of
dep
ress
ion
120 f
t
?
19°
208 y
d
?
1000 y
d
460 f
t
35°
12°
348.5
ft
Lesson 8-5
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
31
G
len
co
e G
eo
me
try
Tw
o A
ng
les
of
Ele
va
tio
n o
r D
ep
ress
ion
A
ngle
s of
ele
vati
on
or
dep
ress
ion
to t
wo
dif
fere
nt
obje
cts
can
be u
sed
to e
stim
ate
dis
tan
ce b
etw
een
th
ose
obje
cts.
Th
e a
ngle
s fr
om
tw
o d
iffe
ren
t p
osi
tion
s of
obse
rvati
on
to t
he s
am
e o
bje
ct c
an
be u
sed
to e
stim
ate
th
e h
eig
ht
of
the o
bje
ct.
T
o e
sti
ma
te t
he
he
igh
t o
f a
ga
ra
ge
,
Ja
so
n s
igh
ts t
he
to
p o
f th
e g
ara
ge
at
a 4
2°
an
gle
of
ele
va
tio
n.
He
th
en
ste
ps b
ack
20
fe
et
an
d s
ite
s
the
to
p a
t a
10
° a
ng
le.
If J
aso
n i
s 6
fe
et
tall
,
ho
w t
all
is t
he
ga
ra
ge
to
th
e n
ea
re
st
foo
t?
△ A
BC
an
d △
AB
D a
re r
igh
t tr
ian
gle
s. W
e c
an
dete
rmin
e A
B =
x a
nd
CB
= y
, an
d D
B =
y +
20.
Use
△ A
BC
. U
se △
AB
D.
tan
42°
= x
−
y o
r y t
an
42°
= x
ta
n 1
0°
=
x
−
y +
20 o
r (y
+ 2
0)
tan
10°
= x
Su
bst
itu
te t
he v
alu
e f
or
x f
rom
△ A
BD
in
th
e e
qu
ati
on
for
△ A
BC
an
d s
olv
e f
or
y.
y t
an
42°
= (
y +
20)
tan
10°
y t
an
42°
= y
tan
10°
+ 2
0 t
an
10°
y t
an
42°
− y
tan
10°
= 2
0 t
an
10°
y (
tan
42°
− t
an
10°)
= 2
0 t
an
10°
y =
20 t
an
10°
−
−
ta
n 4
2°
- t
an
10° ≈
4.8
7
If y
≈ 4
.87,
then
x =
4.8
7 t
an
42°
or
abou
t 4.4
feet.
Ad
d J
aso
n’s
heig
ht,
so t
he g
ara
ge i
s abou
t 4.4
+ 6
or
10.4
feet
tall
.
Exerc
ises
1
. C
LIF
F S
ara
h s
tan
ds
on
th
e g
rou
nd
an
d s
igh
ts t
he t
op
of
a s
teep
cli
ff a
t a 6
0°
an
gle
of
ele
vati
on
. S
he t
hen
st
ep
s back
50 m
ete
rs a
nd
sig
hts
th
e t
op
of
the s
teep
cl
iff
at
a 3
0°
an
gle
. If
Sara
h i
s 1.8
mete
rs t
all
, h
ow
tall
is
the s
teep
cli
ff t
o t
he n
eare
st m
ete
r?
2. B
ALLO
ON
T
he a
ngle
of
dep
ress
ion
fro
m a
hot
air
ball
oon
in
th
e a
ir t
o a
pers
on
on
th
e g
rou
nd
is
36°.
If
the p
ers
on
st
ep
s back
10 f
eet,
th
e n
ew
an
gle
of
dep
ress
ion
is
25°.
If
th
e p
ers
on
is
6 f
eet
tall
, h
ow
far
off
th
e g
rou
nd
is
the
hot
air
ball
oon
?
Stu
dy
Gu
ide a
nd
In
terv
en
tio
n
(con
tin
ued
)
An
gle
s o
f E
levati
on
an
d D
ep
ressio
n
8-5 Exam
ple
42°
10°
x
y6
ft
20 f
t
Bu
ildin
g
60°
30°
x
y1
.8m
50m
Ste
epcl
iff
36°
25°
x
y6
ft
10
ft
Ballo
on
ab
ou
t 45 m
ete
rs
ab
ou
t 19 f
eet
Answers (Lesson 8-5)
Answers
Co
pyri
gh
t ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f T
he
Mc
Gra
w-H
ill C
om
pa
nie
s,
Inc
.
Chapter 8 A15 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
32
G
len
co
e G
eo
me
try
Na
me
th
e a
ng
le o
f d
ep
re
ssio
n o
r a
ng
le o
f e
lev
ati
on
in
ea
ch
fig
ure
.
1
.
2.
3
.
4.
5. M
OU
NT
AIN
BIK
ING
O
n a
mou
nta
in b
ike t
rip
alo
ng t
he G
em
ini
Bri
dges
Tra
il i
n M
oab,
Uta
h,
Nabu
ko s
top
ped
on
th
e c
an
yon
flo
or
to g
et
a g
ood
vie
w o
f th
e t
win
san
dst
on
e
bri
dges.
Nabu
ko i
s st
an
din
g a
bou
t 60 m
ete
rs f
rom
th
e b
ase
of
the c
an
yon
cli
ff,
an
d t
he
natu
ral
arc
h b
rid
ges
are
abou
t 100 m
ete
rs u
p t
he c
an
yon
wall
. If
her
lin
e o
f si
gh
t is
5 m
etr
es
above t
he g
rou
nd
, w
hat
is t
he a
ngle
of
ele
vati
on
to t
he t
op
of
the b
rid
ges?
Rou
nd
to t
he n
eare
st t
en
th d
egre
e.
6. S
HA
DO
WS
S
up
pose
th
e s
un
cast
s a s
had
ow
off
a 3
5-f
oot
bu
ild
ing.
If t
he a
ngle
of
ele
vati
on
to t
he s
un
is
60°,
how
lon
g i
s th
e s
had
ow
to t
he n
eare
st t
en
th o
f a f
oot?
7
. B
ALLO
ON
ING
A
ngie
sees
a h
ot
air
ball
oon
in
th
e
sky f
rom
her
spot
on
th
e g
rou
nd
. T
he a
ngle
of
ele
vati
on
fro
m A
ngie
to t
he b
all
oon
is
40°.
If
she s
tep
s
back
200 f
eet,
th
e n
ew
an
gle
of
ele
vati
on
is
10°.
If A
ngie
is
5.5
feet
tall
, h
ow
far
off
th
e g
rou
nd
is t
he h
ot
air
ball
oon
?
8. IN
DIR
EC
T M
EA
SU
RE
ME
NT
K
yle
is
at
the e
nd
of
a p
ier
30 f
eet
above t
he o
cean
. H
is e
ye l
evel
is
3 f
eet
above t
he p
ier.
He i
s u
sin
g b
inocu
lars
to
watc
h a
wh
ale
su
rface
. If
th
e a
ngle
of
dep
ress
ion
of
the w
hale
is
20°,
how
far
is t
he w
hale
fro
m
Kyle
’s b
inocu
lars
? R
ou
nd
to t
he n
eare
st t
en
th f
oot.
whal
ew
ater
leve
l
20°
Kyl
e’s
eyes
pie
r3 f
t
30 f
t
60° ?
35 f
t
Z
PW
R
D
AC
B
T
WR
S
F
T
L
S
Sk
ills
Practi
ce
An
gle
s o
f E
levati
on
an
d D
ep
ressio
n
8-5
40°
10°
x
y5.5
ft
200 f
t
Bal
loon
ab
ou
t 50.1
ft
ab
ou
t 96.5
ft
ab
ou
t 20.2
ft
ab
ou
t 57.7
°
∠FLS
; ∠
TS
L∠
RTW
; ∠
SW
T
∠D
CB
; ∠
AB
C∠
WZ
P; ∠
RP
Z
Lesson 8-5
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
33
G
len
co
e G
eo
me
try
Na
me
th
e a
ng
le o
f d
ep
re
ssio
n o
r a
ng
le o
f e
lev
ati
on
in
ea
ch
fig
ure
.
1.
2.
3. W
AT
ER
TO
WE
RS
A
stu
den
t ca
n s
ee a
wate
r to
wer
from
th
e c
lose
st p
oin
t of
the s
occ
er
field
at
San
Lobos
Hig
h S
chool.
Th
e e
dge o
f th
e s
occ
er
field
is
abou
t 110 f
eet
from
th
e
wate
r to
wer
an
d t
he w
ate
r to
wer
stan
ds
at
a h
eig
ht
of
32.5
feet.
Wh
at
is t
he a
ngle
of
ele
vati
on
if
the e
ye l
evel
of
the s
tud
en
t vie
win
g t
he t
ow
er
from
th
e e
dge o
f th
e s
occ
er
field
is
6 f
eet
above t
he g
rou
nd
? R
ou
nd
to t
he n
eare
st t
en
th.
4. C
ON
ST
RU
CT
ION
A
roofe
r p
rop
s a l
ad
der
again
st a
wall
so t
hat
the t
op
of
the l
ad
der
reach
es
a 3
0-f
oot
roof
that
need
s re
pair
. If
th
e a
ngle
of
ele
vati
on
fro
m t
he b
ott
om
of
the
lad
der
to t
he r
oof
is 5
5°,
how
far
is t
he l
ad
der
from
th
e b
ase
of
the w
all
? R
ou
nd
you
r
an
swer
to t
he n
eare
st f
oot.
5. T
OW
N O
RD
INA
NC
ES
T
he t
ow
n o
f B
elm
on
t re
stri
cts
the h
eig
ht
of
flagp
ole
s to
25 f
eet
on
an
y p
rop
ert
y.
Lin
dsa
y w
an
ts t
o
dete
rmin
e w
heth
er
her
sch
ool
is i
n c
om
pli
an
ce w
ith
th
e r
egu
lati
on
.
Her
eye l
evel
is 5
.5 f
eet
from
th
e g
rou
nd
an
d s
he s
tan
ds
36 f
eet
from
th
e f
lagp
ole
. If
th
e a
ngle
of
ele
vati
on
is
abou
t 25°,
wh
at
is t
he h
eig
ht
of
the f
lagp
ole
to t
he n
eare
st t
en
th?
6. G
EO
GR
APH
Y S
tep
han
is
stan
din
g o
n t
he g
rou
nd
by
a m
esa
in
th
e P
ain
ted
Dese
rt.
Ste
ph
an
is
1.8
mete
rs
tall
an
d s
igh
ts t
he t
op
of
the m
esa
at
29°.
Ste
ph
an
ste
ps
back
100 m
ete
rs a
nd
sig
hts
th
e t
op
at
25°.
How
tall
is
the m
esa
?
7. IN
DIR
EC
T M
EA
SU
RE
ME
NT
M
r. D
om
ingu
ez i
s st
an
din
g
on
a 4
0-f
oot
oce
an
blu
ff n
ear
his
hom
e.
He c
an
see h
is t
wo
dogs
on
th
e b
each
belo
w.
If h
is l
ine o
f si
gh
t is
6 f
eet
above
the g
rou
nd
an
d t
he a
ngle
s of
dep
ress
ion
to h
is d
ogs
are
34°
an
d 4
8°,
how
far
ap
art
are
th
e d
ogs
to t
he n
eare
st f
oot?
48°
34°
40
ft
6 f
t
Mr.
Do
min
gu
ez
blu
ff
25°
5.5
ft
36
ft
x
R
M
P
L
T
YR
ZY
Practi
ce
An
gle
s o
f E
levati
on
an
d D
ep
ressio
n
8-5
29°
25°
x
y1
.8 m
10
0 m
mes
a
ab
ou
t 21 f
t
ab
ou
t 22.3
ft
ab
ou
t 27 f
t
ab
ou
t 296 m
ab
ou
t 13.5
°
∠TR
Z; ∠
YZ
R∠
PR
M; ∠
LM
R
Answers (Lesson 8-5)
Co
pyrig
ht ©
Gle
nc
oe
/Mc
Gra
w-H
ill, a d
ivis
ion
of T
he
Mc
Gra
w-H
ill Co
mp
an
ies, In
c.
Chapter 8 A16 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
34
G
len
co
e G
eo
me
try
Wo
rd
Pro
ble
m P
racti
ce
An
gle
s o
f E
levati
on
an
d D
ep
ressio
n
1
. LIG
HT
HO
US
ES
S
ail
ors
on
a s
hip
at
sea
spot
the l
igh
t fr
om
a l
igh
thou
se.
Th
e
an
gle
of
ele
vati
on
to t
he l
igh
t is
25°.
25°
30m
?
T
he l
igh
t of
the l
igh
thou
se i
s 30 m
ete
rs
above s
ea l
evel.
How
far
from
th
e s
hore
is t
he s
hip
? R
ou
nd
you
r an
swer
to t
he
neare
st m
ete
r.
2
. R
ES
CU
E A
hik
er
dro
pp
ed
his
back
pack
over
on
e s
ide o
f a c
an
yon
on
to a
led
ge
belo
w.
Beca
use
of
the s
hap
e o
f th
e c
liff
,
he c
ou
ld n
ot
see e
xact
ly w
here
it
lan
ded
.
32°
115 f
t
? f
t
Bac
kpac
k
F
rom
th
e o
ther
sid
e,
the p
ark
ran
ger
rep
ort
s th
at
the a
ngle
of
dep
ress
ion
to
the b
ack
pack
is
32°.
If
the w
idth
of
the
can
yon
is
115 f
eet,
how
far
dow
n d
id
the b
ack
pack
fall
? R
ou
nd
you
r an
swer
to t
he n
eare
st f
oot.
3
. A
IRP
LA
NE
S T
he a
ngle
of
ele
vati
on
to
an
air
pla
ne v
iew
ed
fro
m t
he c
on
trol
tow
er
at
an
air
port
is
7°.
Th
e t
ow
er
is
200 f
eet
hig
h a
nd
th
e p
ilot
rep
ort
s th
at
the a
ltit
ud
e i
s 5200 f
eet.
How
far
aw
ay
from
th
e c
on
trol
tow
er
is t
he a
irp
lan
e?
Rou
nd
you
r an
swer
to t
he n
eare
st f
oot.
4
. P
EA
K T
RA
M T
he P
eak
Tra
m i
n H
on
g
Kon
g c
on
nect
s tw
o t
erm
inals
, on
e a
t th
e
base
of
a m
ou
nta
in,
an
d t
he o
ther
at
the
sum
mit
. T
he a
ngle
of
ele
vati
on
of
the
up
per
term
inal
from
th
e l
ow
er
term
inal
is a
bou
t 15.5
°. T
he d
ista
nce
betw
een
th
e
two t
erm
inals
is
abou
t 1365 m
ete
rs.
Abou
t h
ow
mu
ch h
igh
er
above s
ea l
evel
is t
he u
pp
er
term
inal
com
pare
d t
o t
he
low
er
term
inal?
Rou
nd
you
r an
swer
to
the n
eare
st m
ete
r.
5
. H
ELIC
OP
TE
RS
Jerm
ain
e a
nd
Joh
n a
re
watc
hin
g a
heli
cop
ter
hover
above t
he
gro
un
d.
48°
55°
(Not
dra
wn t
o s
cale
)
Hel
icopte
r
Jerm
aine
John
10 m
h
x
Jerm
ain
e a
nd
Joh
n a
re s
tan
din
g
10 m
ete
rs a
part
.
a.
Fin
d t
wo d
iffe
ren
t exp
ress
ion
s th
at
can
be u
sed
to f
ind
th
e h
, h
eig
ht
of
the h
eli
cop
ter.
b.
Equ
ate
th
e t
wo e
xp
ress
ion
s you
fou
nd
for
Exerc
ise a
to s
olv
e f
or x.
Rou
nd
you
r an
swer
to t
he n
eare
st
hu
nd
red
th.
c.
How
hig
h a
bove t
he g
rou
nd
is
the
heli
cop
ter?
Rou
nd
you
r an
swer
to t
he
neare
st h
un
dre
dth
.
8-5 6
4 m
72 f
t
41,0
28 f
t
365 m
h =
x t
an
55°;
h
= (
x +
10)
tan
48°
34.9
8 m
49.9
5 m
Lesson 8-5
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
35
G
len
co
e G
eo
me
try
En
ric
hm
en
t
Best
Seat
in t
he H
ou
se
Most
peop
le w
an
t to
sit
in
th
e b
est
seat
in t
he m
ovie
th
eate
r.
Th
e b
est
seat
cou
ld b
e
defi
ned
as
the s
eat
that
all
ow
s you
to s
ee t
he m
axim
um
am
ou
nt
of
scre
en
. T
he p
ictu
re
belo
w r
ep
rese
nts
th
is s
itu
ati
on
.
To d
ete
rmin
e t
he b
est
seat
in t
he h
ou
se,
you
wan
t to
fin
d w
hat
valu
e o
f x a
llow
s you
to s
ee
the m
axim
um
am
ou
nt
of
scre
en
. T
he v
alu
e o
f x i
s h
ow
far
from
th
e s
creen
you
sh
ou
ld s
it.
1
. T
o m
axim
ize t
he a
mou
nt
of
scre
en
vie
wed
, w
hic
h a
ngle
valu
e n
eed
s to
be m
axim
ized
?
Wh
y?
2
. W
hat
is t
he v
alu
e o
f a
if x =
10 f
eet?
3
. W
hat
is t
he v
alu
e o
f a
if x =
20 f
eet?
4
. W
hat
is t
he v
alu
e o
f a
if x =
25 f
eet?
5
. W
hat
is t
he v
alu
e o
f a
if x =
35 f
eet?
6
. W
hat
is t
he v
alu
e o
f a
if x =
55 f
eet?
7. W
hic
h v
alu
e o
f x g
ives
the g
reate
st v
alu
e o
f a
? S
o,
wh
ere
is
the b
est
seat
in t
he m
ovie
theate
r?
a ˚
b˚
c ˚ xft
scre
en
40
ft
10
ft
8-5 A
ng
le w
ith
measu
re a
becau
se t
his
is h
ow
mu
ch
of
the s
cre
en
can
b
e v
iew
ed
.
33.7
ft
41.6
ft
41.6
ft
39.1
ft
32 f
t
Th
e b
est
seat
in t
he h
ou
se i
s a
rou
nd
20–25 f
eet
aw
ay f
rom
th
e s
cre
en
.
Answers (Lesson 8-5)
Answers
Co
pyri
gh
t ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f T
he
Mc
Gra
w-H
ill C
om
pa
nie
s,
Inc
.
Chapter 8 A17 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
36
G
len
co
e G
eo
me
try
Stu
dy
Gu
ide a
nd
In
terv
en
tio
n
Th
e L
aw
of
Sin
es a
nd
Law
of
Co
sin
es
Th
e L
aw
of S
ine
s
In a
ny t
rian
gle
, th
ere
is
a s
peci
al
rela
tion
ship
betw
een
th
e a
ngle
s of
the t
rian
gle
an
d t
he l
en
gth
s of
the s
ides
op
posi
te t
he a
ngle
s.
F
ind
b.
Ro
un
d t
o t
he
ne
are
st
ten
th.
45
°
30
74
°
b
B
AC
sin
C
−
c
= si
n B
−
b
Law
of
Sin
es
si
n 4
5°
−
30
=
sin
74°
−
b
m
∠C
= 4
5,
c =
30,
m∠
B =
74
b s
in 4
5°
= 3
0 s
in 7
4°
Cro
ss P
roducts
Pro
pert
y
b =
30 s
in 7
4°
−
sin
45
°
Div
ide e
ach s
ide b
y s
in 4
5°.
b ≈
40.8
U
se a
calc
ula
tor.
F
ind
d.
Ro
un
d t
o t
he
ne
are
st
ten
th.
By t
he T
rian
gle
An
gle
-Su
m T
heore
m,
m∠
E =
180 -
(82 +
40)
or
58.
82
°40
°
24
d
si
n D
−
d
= si
n E
−
e
Law
of
Sin
es
si
n 8
2°
−
d
= si
n 5
8°
−
24
m
∠D
= 8
2,
m∠
E =
58,
e =
24
24 s
in 8
2°
= d
sin
58°
Cro
ss P
roducts
Pro
pert
y
24 s
in 8
2°
−
sin
58°
= d
D
ivid
e e
ach s
ide b
y s
in 5
8°.
d
≈ 2
8.0
U
se a
calc
ula
tor.
Exerc
ises
Fin
d x
. R
ou
nd
to
th
e n
ea
re
st
ten
th.
1.
80°
40°
x
12
2
.
52°
91°
x
20
3
.
16
x
84°
34°
4
.
17°
72°
x25
5
.
89°
80°
x12
6.
60°
35°
35
x
8-6
Exam
ple
1Exam
ple
2
La
w o
f S
ine
s s
in A
−
a
= s
in B
−
b
= s
in C
−
c
7.7
12.2
30.4
18.4
26.2
18.0
Lesson 8-6
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
37
G
len
co
e G
eo
me
try
Stu
dy
Gu
ide a
nd
In
terv
en
tio
n
(con
tin
ued
)
Th
e L
aw
of
Sin
es a
nd
Law
of
Co
sin
es
Th
e L
aw
of C
osin
es
An
oth
er
rela
tion
ship
betw
een
th
e s
ides
an
d a
ngle
s of
an
y t
rian
gle
is
call
ed
th
e L
aw
of
Co
sin
es.
You
can
use
th
e L
aw
of
Cosi
nes
if y
ou
kn
ow
th
ree s
ides
of
a
tria
ngle
or
if y
ou
kn
ow
tw
o s
ides
an
d t
he i
ncl
ud
ed
an
gle
of
a t
rian
gle
.
La
w o
f C
os
ine
s
Le
t △
AB
C b
e a
ny t
ria
ng
le w
ith
a,
b,
an
d c
re
pre
se
ntin
g t
he
me
asu
res o
f th
e s
ide
s o
pp
osite
the
an
gle
s w
ith
me
asu
res A
, B
, a
nd
C,
resp
ective
ly.
Th
en
th
e f
ollo
win
g e
qu
atio
ns a
re t
rue
.
a
2 =
b2 +
c2 -
2b
c c
os A
b
2 =
a2 -
c2 -
2a
c c
os B
c
2 =
a2 +
b2 -
2a
b c
os C
F
ind
c.
Ro
un
d t
o t
he
ne
are
st
ten
th.
c
2 =
a2 +
b2 -
2a
b c
os
C
Law
of
Cosin
es
c
2 =
12
2 +
10
2 -
2(1
2)(
10)c
os
48
° a
= 1
2,
b =
10,
m∠
C =
48
c =
√ '
''
''
''
''
''
'
12
2 +
10
2 -
2(1
2)(
10)c
os
48
° T
ake t
he s
quare
root
of
each s
ide.
c ≈
9.1
U
se a
calc
ula
tor.
F
ind
m∠A
. R
ou
nd
to
th
e n
ea
re
st
de
gre
e.
a
2 =
b2 +
c2 -
2bc c
os
A
Law
of
Cosin
es
7
2 =
52 +
82 -
2(5
)(8)
cos
A
a =
7,
b =
5,
c =
8
49 =
25 +
64 -
80 c
os
A
Multip
ly.
-
40 =
-80 c
os
A
Subtr
act
89 f
rom
each s
ide.
1
−
2 =
cos
A
Div
ide e
ach s
ide b
y -
80.
cos-
1 1
−
2 =
A
Use t
he invers
e c
osin
e.
60°
= A
U
se a
calc
ula
tor.
Exerc
ises
Fin
d x
. R
ou
nd
an
gle
me
asu
re
s t
o t
he
ne
are
st
de
gre
e a
nd
sid
e m
ea
su
re
s t
o t
he
ne
are
st
ten
th.
1
.
2.
3
.
4
.
5.
6
.
8-6
Exam
ple
1
Exam
ple
2
48
°10
12
c
C
BA
12
14
x
62°
12
11
10
x°
16
24
18x°
25
20
x
82°
18
28
x
59°
18
15
15
x°
8
5
7
C
B
A
13.5
51
42
29.8
24.3
53
Answers (Lesson 8-6)
Co
pyrig
ht ©
Gle
nc
oe
/Mc
Gra
w-H
ill, a d
ivis
ion
of T
he
Mc
Gra
w-H
ill Co
mp
an
ies, In
c.
Chapter 8 A18 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
38
G
len
co
e G
eo
me
try
Fin
d x
. R
ou
nd
an
gle
me
asu
re
s t
o t
he
ne
are
st
de
gre
e a
nd
sid
e l
en
gth
s t
o t
he
ne
are
st
ten
th.
1.
28
35°
48°
x
2.
18
46°
17°
x
3.
38
x
86°
51°
4
.
5.
6
.
7
.
8.
9
.
10
.
11
.
12
.
13
.
14
.
15
.
16
. S
olv
e t
he t
rian
gle
. R
ou
nd
an
gle
measu
res
to t
he n
eare
st d
egre
e.
Sk
ills
Practi
ce
Th
e L
aw
of
Sin
es a
nd
Law
of
Co
sin
es
8-6
x8
73°80°
x
34
41°
88°
x
12
83°
60°
18
x
104°
22°
x27
54°
93°
12
16
11
x°
17
18
23
29
111°
34x
16
89°
12x
13
103°
20
x
13
35
x
96°
28 x°
26
30
25
28°
x20
21.6
7.3
48.8
46.6
40.1
88
52
19.8
26.2
16.8
19.8
43.7
38.6
11.9
67
m∠
A =
82,
m∠
B =
51,
m∠
C =
47
Lesson 8-6
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
39
G
len
co
e G
eo
me
try
Practi
ce
Th
e L
aw
of
Sin
es a
nd
Law
of
Co
sin
es
13. IN
DIR
EC
T M
EA
SU
RE
ME
NT
T
o f
ind
th
e d
ista
nce
fro
m t
he e
dge
of
the l
ak
e t
o t
he t
ree o
n t
he i
slan
d i
n t
he l
ak
e,
Han
nah
set
up
a t
rian
gu
lar
con
figu
rati
on
as
show
n i
n t
he d
iagra
m.
Th
e d
ista
nce
from
loca
tion
A t
o l
oca
tion
B i
s 85 m
ete
rs.
Th
e m
easu
res
of
the
an
gle
s at A
an
d B
are
51°
an
d 8
3°,
resp
ect
ively
. W
hat
is t
he
dis
tan
ce f
rom
th
e e
dge o
f th
e l
ak
e a
t B
to t
he t
ree o
n t
he i
slan
d a
t C
?
A
C
B
Fin
d x
. R
ou
nd
an
gle
me
asu
re
s t
o t
he
ne
are
st
de
gre
e a
nd
sid
e l
en
gth
s t
o t
he
ne
are
st
ten
th.
1
.
2.
3
.
4
.
5.
6
.
7
.
8.
9
.
10
.
11
.
12
.
14
x
67°
14°
127
x
42°
61°
x
5.8
83°
73°
19.1
x
56°
34°
9.6
x
43° 123°
4.3
x
85°
64°
40
12
37°
x23
28
37°
x
15
17
62°
x
28.4
14.5
21.7
x°
89°
14
15
x
14
10
9.6
x°
8-6
3.7
166,
014.2
31.3
16.9
16.6
48
20.3
43
12.9
8.3
33.2
ab
ou
t 91.8
m
Answers (Lesson 8-6)
Answers
Co
pyri
gh
t ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f T
he
Mc
Gra
w-H
ill C
om
pa
nie
s,
Inc
.
Chapter 8 A19 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
40
G
len
co
e G
eo
me
try
Wo
rd
Pro
ble
m P
racti
ce
Th
e L
aw
of
Sin
es a
nd
Law
of
Co
sin
es
1
. A
LT
ITU
DE
S In
tri
an
gle
AB
C,
the
alt
itu
de t
o s
ide −
−
AB
is
dra
wn
.
A
ab
B
C
G
ive t
wo e
xp
ress
ion
s fo
r th
e l
en
gth
of
the a
ltit
ud
e i
n t
erm
s of
a,
b,
an
d t
he
sin
e o
f th
e a
ngle
s A
an
d B
.
2
. M
AP
S T
hre
e c
itie
s fo
rm t
he v
ert
ices
of
a t
rian
gle
. T
he a
ngle
s of
the t
rian
gle
are
40°,
60°,
an
d 8
0°.
Th
e t
wo m
ost
dis
tan
t ci
ties
are
40 m
iles
ap
art
. H
ow
clo
se a
re
the t
wo c
lose
st c
itie
s? R
ou
nd
you
r an
swer
to t
he n
eare
st t
en
th o
f a m
ile.
3
. S
TA
TU
ES
G
ail
was
vis
itin
g a
n a
rt
gall
ery
. In
on
e r
oom
, sh
e s
tood
so t
hat
she h
ad
a v
iew
of
two s
tatu
es,
on
e o
f a
man
, an
d t
he o
ther
of
a w
om
an
. S
he w
as
40 f
eet
from
th
e s
tatu
e o
f th
e w
om
an
, an
d 3
5 f
eet
from
th
e s
tatu
e o
f th
e m
an
. T
he a
ngle
cre
ate
d b
y t
he l
ines
of
sigh
t to
th
e t
wo s
tatu
es
was
21°.
Wh
at
is t
he
dis
tan
ce b
etw
een
th
e t
wo s
tatu
es?
R
ou
nd
you
r an
swer
to t
he n
eare
st t
en
th.
4
. C
AR
S T
wo c
ars
sta
rt m
ovin
g f
rom
th
e
sam
e l
oca
tion
. T
hey h
ead
str
aig
ht,
bu
t in
dif
fere
nt
dir
ect
ion
s. T
he a
ngle
betw
een
wh
ere
th
ey a
re h
ead
ing i
s 43
°.
Th
e f
irst
car
travels
20 m
iles
an
d t
he
seco
nd
car
travels
37 m
iles.
How
far
ap
art
are
th
e t
wo c
ars
? R
ou
nd
you
r an
swer
to t
he n
eare
st t
en
th.
5
. IS
LA
ND
S O
ah
u
is a
Haw
aii
an
Is
lan
d.
Off
of
the
coast
of
Oah
u,
there
is
a v
ery
ti
ny i
slan
d
kn
ow
n a
s C
hin
am
an
’s H
at.
K
eok
i an
d M
ali
a
are
obse
rvin
g
Ch
inam
an
’s H
at
from
loca
tion
s 5 k
ilom
ete
rs a
part
. U
se t
he i
nfo
rmati
on
in
th
e f
igu
re t
o a
nsw
er
the f
oll
ow
ing
qu
est
ion
s.
a.
How
far
is K
eok
i fr
om
Ch
inam
an
’s
Hat?
Rou
nd
you
r an
swer
to t
he
neare
st t
en
th o
f a k
ilom
ete
r.
b.
How
far
is M
ali
a f
rom
Ch
inam
an
’s
Hat?
Rou
nd
you
r an
swer
to t
he
neare
st t
en
th o
f a k
ilom
ete
r.
Ka’a
’aw
a
Keoki
Chin
am
an’s
Hat
5 k
m
Malia
Kahalu
’u
22.3
°
49.5
°
8-6
Sta
tue o
f
a m
an
Sta
tue o
f a w
om
an
35 f
t
40 f
t
21˚
a s
in B
= b
sin
A =
len
gth
of
alt
itu
de
26.1
mi
14.5
ft
26.2
mi
4.0
km
2.0
km
Lesson 8-6
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
41
G
len
co
e G
eo
me
try
En
ric
hm
en
t
Iden
titi
es
An
id
en
tity
is
an
equ
ati
on
th
at
is t
rue f
or
all
valu
es
of
the
vari
able
for
wh
ich
both
sid
es
are
defi
ned
. O
ne w
ay t
o v
eri
fy
an
id
en
tity
is
to u
se a
rig
ht
tria
ngle
an
d t
he d
efi
nit
ion
s fo
r tr
igon
om
etr
ic f
un
ctio
ns.
V
erif
y t
ha
t (s
in A
)2 +
(co
s A
)2 =
1 i
s a
n i
de
nti
ty.
(si
n A
)2 +
(co
s A
)2 =
( a
−
c ) 2
+ ( b
−
c ) 2
=
a2+
b2
−
c2
=
c2
−
c2
= 1
To c
heck
wh
eth
er
an
equ
ati
on
ma
y b
e a
n i
den
tity
, you
can
test
se
vera
l valu
es.
How
ever,
sin
ce y
ou
can
not
test
all
valu
es,
you
ca
nn
ot
be c
erta
in t
hat
the e
qu
ati
on
is
an
id
en
tity
.
T
est
sin
2x =
2 s
in x
co
s x
to
se
e i
f it
co
uld
be
an
id
en
tity
.
Try
x =
20.
Use
a c
alc
ula
tor
to e
valu
ate
each
exp
ress
ion
.
sin
2x =
sin
40
2 s
in x
cos
x =
2 (
sin
20)(
cos
20)
≈
0.6
43
≈ 2
(0.3
42)(
0.9
40)
≈
0.6
43
Sin
ce t
he l
eft
an
d r
igh
t si
des
seem
equ
al,
th
e e
qu
ati
on
may b
e a
n i
den
tity
.
Exerc
ises
Use
tria
ng
le ABC
sh
ow
n a
bo
ve
. V
erif
y t
ha
t e
ach
eq
ua
tio
n i
s a
n i
de
nti
ty.
1
. co
s A
−
sin
A =
1
−
tan
A
2. ta
n B
−
sin
B =
1
−
cos
B
3. ta
n B
cos
B =
sin
B
4. 1 -
(co
s B
)2 =
(si
n B
)2
Try
se
ve
ra
l v
alu
es f
or x
to
te
st
wh
eth
er e
ach
eq
ua
tio
n c
ou
ld b
e a
n i
de
nti
ty.
5
. co
s 2x =
(co
s x)2
- (
sin
x)2
6. co
s (9
0 -
x)
= s
in x
B
AC
ca
b
8-6
Exam
ple
1
Exam
ple
2
tan
B
−
sin
B =
b
−
a
÷ b
−
c =
c
−
a =
1
−
co
s B
c
os
A
−
sin
A =
b
−
c
÷ a
−
c =
b
−
a =
1
−
tan
A
tan
B c
os B
= b
−
a
· a
−
c =
b
−
c =
sin
B1 -
(co
s B
)2 =
1 -
( a
−
c ) 2
=
c2
−
c
2 -
a
2
−
c
2
=
c2 -
a2
−
c
2
= b
2
−
c
2 o
r (s
in B
)2
Yes;
see s
tud
en
ts’
wo
rk.
Yes;
see s
tud
en
ts’
wo
rk.
Answers (Lesson 8-6)
Co
pyrig
ht ©
Gle
nc
oe
/Mc
Gra
w-H
ill, a d
ivis
ion
of T
he
Mc
Gra
w-H
ill Co
mp
an
ies, In
c.
Chapter 8 A20 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
42
G
len
co
e G
eo
me
try
Gra
ph
ing C
alc
ula
tor
Act
ivit
y
So
lvin
g T
rian
gle
s U
sin
g t
he L
aw
of
Sin
es o
r C
osin
es
You
can
use
a c
alc
ula
tor
to s
olv
e t
rian
gle
s u
sin
g t
he L
aw
of
Sin
es
or
Cosi
nes.
S
olv
e △ABC i
f a
= 6
, b
= 2
, a
nd
c =
7.5
.
U
se t
he L
aw
of
Cosi
nes.
a
2 =
b2 +
c2 -
2bc c
os A
6
2 =
22 +
7.5
2 -
2(2
)(7.5
) co
s A
m
∠A =
cos-
1 6
2 -
22 -
7.5
2
−
-2(2
)(7.5
)
Use
you
r ca
lcu
lato
r to
fin
d t
he m
easu
re o
f ∠A.
K
ey
str
ok
es:
2nd
[C
OS
-1]
( 6
x
2
— 2
x
2
— 7
.5
x2
)
÷
En
ter:
(
(–) 2
3
2
3 7
.5
)
)
EN
TE
R
36.06658826
S
o m
∠A
≈ 3
6.
Use
th
e L
aw
of
Sin
es
an
d y
ou
r ca
lcu
lato
r to
fin
d m
∠B.
si
n A
−
a =
sin
B
−
b
si
n 3
6
−
6
≈
si
n B
−
2
m
∠B ≈
sin
-1 2
sin
36°
−
6
K
ey
str
ok
es:
2nd
[S
IN-
1]
( 2
S
IN 3
6
)
÷ 6
)
E
NT
ER
11.29896425
So m
∠B
≈ 1
1.
By t
he T
rian
gle
An
gle
-Su
m T
heore
m, m
∠C ≈
180 -
(36 +
11)
or
133.
Exerc
ises
So
lve
ea
ch
△ABC
. R
ou
nd
me
asu
re
s o
f sid
es t
o t
he
ne
are
st
ten
th
an
d m
ea
su
re
s o
f a
ng
les t
o t
he
ne
are
st
de
gre
e.
1
. a
= 9
, b
= 1
4, c =
12
2
. m
∠C
= 8
0, c =
9, m
∠A
= 4
0
3
. m
∠B
= 4
5, m
∠C
= 5
6, a
= 2
4
. a =
5.7
, b =
6, c =
5
5
. a =
11, b =
15, c =
21
8-6 Exam
ple
m∠
A ≈
40,
m∠
B ≈
89,
m∠
C ≈
51
m∠
B =
60,
a ≈
5.9
, b
≈ 7
.9
m∠
A =
79,
b ≈
1.4
, c ≈
1.7
m∠
A =
62,
m∠
B =
68,
m∠
C =
50
m∠
A ≈
30,
m∠
B ≈
43,
m∠
C ≈
107
Lesson 8-7
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
43
G
len
co
e G
eo
me
try
Geo
metr
ic V
ect
or
Op
era
tio
ns
A v
ect
or
is a
dir
ect
ed s
egm
en
t re
pre
sen
tin
g a
qu
an
tity
th
at
has
both
ma
gn
itu
de, or
len
gth
, an
d d
irecti
on
. F
or
exam
ple
, th
e s
peed a
nd d
irect
ion
of
an
air
pla
ne c
an
be r
epre
sen
ted b
y a
vect
or.
In
sym
bols
, a v
ect
or
is w
ritt
en
as
'
AB ,
wh
ere
A i
s th
e
init
ial
poin
t an
d B
is
the e
ndpoin
t, o
r as
'
v .
Th
e s
um
of
two v
ect
ors
is
call
ed t
he r
esu
lta
nt.
S
ubtr
act
ing a
vect
or
is e
qu
ivale
nt
to a
ddin
g i
ts o
pposi
te. T
he r
esu
ltan
t of
two v
ect
ors
can
be
fou
nd u
sin
g t
he p
ara
llelo
gra
m m
eth
od
or
the t
ria
ng
le m
eth
od
.
C
op
y t
he
ve
cto
rs t
o f
ind
$ a -
$ b .
a(
b(
Me
tho
d 1
: U
se t
he p
ara
llelo
gra
m m
eth
od
.
Co
py ' a
an
d -
'
b w
ith
th
e s
am
e in
itia
l
po
int.
a
-b
Co
mp
lete
th
e p
ara
llelo
gra
m.
a
-b
Dra
w t
he
dia
go
na
l o
f th
e
pa
ralle
log
ram
fro
m t
he
in
itia
l p
oin
t.
a
a
-b
-b
Me
tho
d 2
: U
se t
he t
rian
gle
meth
od
.
Co
py ' a
.
a
Pla
ce
th
e in
itia
l p
oin
t o
f -
'
b a
t th
e
term
ina
l p
oin
t o
f ' a .
a-b
Dra
w t
he
ve
cto
r fr
om
th
e in
itia
l
po
int
of
'
a t
o t
he
te
rmin
al p
oin
t o
f
- '
b
.
a -b
a -
b
Exerc
ises
Co
py
th
e v
ecto
rs.
Th
en
fin
d e
ach
su
m o
r d
iffe
re
nce
.
1
. ' c
+ '
d
2. w
- '
z
3. ' a
- '
b
4. ' r
+ '
t
((
Exam
ple
c(d(
c(d(
+c(
d(
w(
z(
w(
w(z(
-( z
r(
t(
r(t(
+
r(
t(
a(
b(
a( -b(
a(
b(
Stu
dy
Gu
ide a
nd
In
terv
en
tio
n
Vecto
rs
8-7
Answers (Lesson 8-6 and Lesson 8-7)
Answers
Co
pyri
gh
t ©
Gle
nco
e/M
cG
raw
-Hill, a
div
isio
n o
f T
he
Mc
Gra
w-H
ill C
om
pa
nie
s,
Inc
.
Chapter 8 A21 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
44
G
len
co
e G
eo
me
try
Stu
dy
Gu
ide a
nd
In
terv
en
tio
n
(conti
nued
)
Vectors
8-7
Ve
cto
rs o
n t
he
Co
ord
ina
te
Pla
ne
A
vect
or
in s
tan
da
rd
po
sit
ion
h
as
its
init
ial
poin
t at
(0,
0)
an
d c
an
be r
ep
rese
nte
d b
y t
he o
rdere
d
pair
for
poin
t B
. T
he v
ect
or
at
the r
igh
t ca
n b
e e
xp
ress
ed
as
� v =
⟨5,
3⟩.
You
can
use
th
e D
ista
nce
Form
ula
to f
ind
th
e m
agn
itu
de
| �
AB
| of
a v
ect
or.
You
can
desc
ribe t
he d
irect
ion
of
a v
ect
or
by m
easu
rin
g t
he a
ngle
th
at
the v
ect
or
form
s w
ith
th
e p
osi
tive
x-a
xis
or
wit
h a
ny o
ther
hori
zon
tal
lin
e.
F
ind
th
e m
ag
nit
ud
e a
nd
dir
ecti
on
of
a =
⟨3
, 5
⟩.
Fin
d t
he m
agn
itu
de.
� a
= √
%%
%%
%%
%%
%
(x
2 -
x1)2
+ (y
2 -
y1)2
=
√ %
%%
%%
%%
%
(3
- 0
)2 +
(5
- 0
)2
=
√ %%
34 o
r abou
t 5.8
To f
ind
th
e d
irect
ion
, u
se t
he t
an
gen
t ra
tio.
ta
n θ
= 5
−
3
The t
angent
ratio is o
pposite o
ver
adja
cent.
m
∠θ ≈
59.0
U
se a
calc
ula
tor.
Th
e m
agn
itu
de o
f th
e v
ect
or
is a
bou
t 5.8
un
its
an
d i
ts d
irect
ion
is
59°.
Exerc
ises
Fin
d t
he
ma
gn
itu
de
an
d d
ire
cti
on
of
ea
ch
ve
cto
r.
1. � b
= ⟨
-5, 2
⟩ 2.
� c =
⟨-
2, 1
⟩
3. � d =
⟨3, 4
⟩ 4.
� m =
⟨5,
-1
⟩
5. � r
= ⟨
-3,
-4
⟩ 6.
� v =
⟨-
4, 1
⟩
x
y
( 3, 5)
x
y
( 5, 3)
( 0, 0)
Exam
ple
5.4
; 158.2
°2.2
; 153.4
°
5;
53.1
°5.1
; 348.7
°
5;
233.1
°4.1
; 166.0
°
Dis
tance F
orm
ula
(x1,
y1)
= (
0,
0)
and (
x2,
y2)
= (
3,
5)
Sim
plif
y.
Lesson 8-7
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
45
G
len
co
e G
eo
me
try
Sk
ills
Pra
ctic
e
Vectors
8-7
Use
a r
ule
r a
nd
a p
ro
tra
cto
r t
o d
ra
w e
ach
ve
cto
r.
Inclu
de
a s
ca
le o
n e
ach
dia
gra
m.
1.
� a =
20 m
ete
rs p
er
seco
nd
2.
� b =
10 p
ou
nd
of
forc
e a
t
60°
west
of
sou
th
135°
to t
he h
ori
zon
tal
Co
py
th
e v
ecto
rs t
o f
ind
ea
ch
su
m o
r d
iffe
re
nce
.
3
. � a
+ � z
4.
� t -
� r
Writ
e t
he
co
mp
on
en
t fo
rm
of
ea
ch
ve
cto
r.
5
.
6.
Fin
d t
he
ma
gn
itu
de
an
d d
ire
cti
on
of
ea
ch
ve
cto
r.
7.
� m =
⟨2,
12
⟩ 8
. � p
= ⟨
3,
10
⟩
9.
� k =
⟨-
8,
-3
⟩ 1
0.
� f =
⟨-
5,
11
⟩
Fin
d e
ach
of
the
fo
llo
win
g f
or
a
= ⟨
2,
4⟩,
b =
⟨3
, -
3⟩
, a
nd
c =
⟨4
, -
1⟩.
Ch
eck
yo
ur
an
sw
ers g
ra
ph
ica
lly
.
x
y
O
B( –2, 2)
D( 3, –3)
x
y
O
C( 1, –1)
E( 4, 3)
a
z
r t
-r
r-
t
t a
z
az
+
N
E
S
W
1 cm : 5 m
60°
a
1 in : 10 lb135°
b
Overm
att
er
2 √
''
37 ≈
12.2
, 80.5
° √
''
109 ≈
10.4
, 73.3
°
√ ''
73 ≈
8.5
, 200.6
° √
''
146 ≈
12.1
, 155.6
°
⟨3,
4⟩
⟨5,
-5
⟩
Answers (Lesson 8-7)
Co
pyrig
ht ©
Gle
nc
oe
/Mc
Gra
w-H
ill, a d
ivis
ion
of T
he
Mc
Gra
w-H
ill Co
mp
an
ies, In
c.
Chapter 8 A22 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
46
G
len
co
e G
eo
me
try
Practi
ce
Vectors
Use
a r
ule
r a
nd
a p
ro
tra
cto
r t
o d
ra
w e
ach
ve
cto
r.
Inclu
de
a s
ca
le o
n e
ach
dia
gra
m.
1.
� v =
12 N
ew
ton
s of
forc
e a
t 2. � w =
15 m
iles
per
hou
r
40°
to t
he h
ori
zon
tal
70°
east
of
nort
h
Co
py
th
e v
ecto
rs t
o f
ind
ea
ch
su
m o
r d
iffe
re
nce
.
3
. � p
+ � r
4. � a
- � b
5. W
rite
th
e c
om
pon
en
t fo
rm o
f
�
AB .
Fin
d t
he
ma
gn
itu
de
an
d d
ire
cti
on
of
ea
ch
ve
cto
r.
6
. � t
= ⟨
6,
11
⟩ 7
. � g
= ⟨
9,
-7
⟩
Fin
d e
ach
of
the
fo
llo
win
g f
or
a
= ⟨
-1
.5,
4⟩,
b =
⟨7
, 3
⟩, a
nd
c =
⟨1
, -
2⟩.
Ch
eck
yo
ur
an
sw
ers g
ra
ph
ica
lly
.
8.
2 � a
+ � b
9. 2 � c
- � b
10.
AV
IAT
ION
A
jet
begin
s a f
ligh
t alo
ng a
path
du
e n
ort
h a
t 300 m
iles
per
hou
r. A
win
d
is b
low
ing d
ue w
est
at
30 m
iles
per
hou
r.
a.
Fin
d t
he r
esu
ltan
t velo
city
of
the p
lan
e.
b.
Fin
d t
he r
esu
ltan
t d
irect
ion
of
the p
lan
e.
8-7
x
y
O
K( –2, 4)
L( 3, –4)
p%r%
p
r p
+r
a%b%
-b
a-
b
a
1 cm : 4 N
40°
V
N
ES
W
1 in : 10 m
i
70°
w
⟨5,
-8
⟩
ab
ou
t 301.5
mp
h
ab
ou
t 5.7
° west
of
du
e n
ort
h
b 2
a
x
y
−448
12
4−4
812
2a
+ b
b
y
−4
−848
x
−8
−12
4
2c
-b
2c
⟨-1,
13
⟩⟨-
5,
-1
⟩
Lesson 8-7
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
47
G
len
co
e G
eo
me
try
Wo
rd
Pro
ble
m P
racti
ce
Vectors
8-7
1
. W
IND
T
he v
ect
or
�v re
pre
sen
ts t
he s
peed
an
d d
irect
ion
th
at
the w
ind
is
blo
win
g.
Su
dd
en
ly t
he w
ind
pic
ks
up
an
d d
ou
ble
s
its
speed
, bu
t th
e d
irect
ion
does
not
chan
ge.
Wri
te a
n e
xp
ress
ion
for
a v
ect
or
that
desc
ribes
the n
ew
win
d v
elo
city
in
term
s of
�v.
2
. SW
IMM
ING
Jan
is
swim
min
g i
n a
tria
thlo
n e
ven
t. W
hen
th
e o
cean
wate
r
is s
till
, h
er
velo
city
can
be r
ep
rese
nte
d
by t
he v
ect
or
⟨2,
1⟩
mil
es
per
hou
r.
Du
rin
g t
he c
om
peti
tion
, th
ere
was
a
fierc
e c
urr
en
t re
pre
sen
ted
by t
he
vect
or
⟨–1,
–1
⟩ m
iles
per
hou
r. W
hat
vect
or
rep
rese
nts
Jan
’s v
elo
city
du
rin
g
the r
ace
?
3
. PO
LY
GO
NS
D
raw
a r
egu
lar
poly
gon
aro
un
d t
he o
rigin
. F
or
each
sid
e o
f th
e
poly
gon
, ass
oci
ate
a v
ect
or
wh
ose
magn
itu
de i
s th
e l
en
gth
of
the
corr
esp
on
din
g s
ide a
nd
wh
ose
dir
ect
ion
poin
ts i
n t
he c
lock
wis
e m
oti
on
aro
un
d
the o
rigin
. W
hat
vect
or
rep
rese
nts
th
e
sum
of
all
th
ese
vect
ors
? E
xp
lain
.
4
. B
AS
EB
ALL R
ick
is
in t
he m
idd
le o
f a
base
ball
gam
e.
His
team
mate
th
row
s
him
th
e b
all
, bu
t th
row
s it
far
in f
ron
t of
him
. H
e h
as
to r
un
as
fast
as
he c
an
to
catc
h i
t. A
s h
e r
un
s, h
e k
now
s th
at
as
soon
as
he c
atc
hes
it,
he h
as
to t
hro
w i
t
as
hard
as
he c
an
to t
he t
eam
mate
at
hom
e p
late
. H
e h
as
no t
ime t
o s
top
. In
the f
igu
re,
�x is
th
e v
ect
or
that
rep
rese
nts
the v
elo
city
of
the b
all
after
Ric
k t
hro
ws
it a
nd
�v
rep
rese
nts
Ric
k’s
velo
city
beca
use
he i
s ru
nn
ing.
Ass
um
e t
hat
Ric
k
can
th
row
ju
st a
s h
ard
wh
en
ru
nn
ing a
s
he c
an
wh
en
sta
nd
ing s
till
. Homeplate
Rick
x
v%
%% X
% V
a.
Wh
at
vect
or
wou
ld r
ep
rese
nt
the
velo
city
of
the b
all
if
Ric
k t
hre
w i
t th
e
sam
e w
ay b
ut
he w
as
stan
din
g s
till
?
b.
Th
e a
ngle
betw
een
� x an
d � v
is 8
9°.
By
run
nin
g,
did
it
help
Ric
k g
et
the b
all
to h
om
e p
late
fast
er
than
he w
ou
ld
have n
orm
all
y b
een
able
to i
f h
e w
ere
stan
din
g s
till
?
2
v
⟨1,
0⟩
mp
h
x -
v
⟨0,
0⟩;
Becau
se t
he p
oly
go
n
form
s a
clo
sed
lo
op
, th
e s
um
is
th
e z
ero
vecto
r.
Yes,
it h
elp
ed
.
Answers (Lesson 8-7)
Answers
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.
Chapter 8 A23 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
8
48
G
len
co
e G
eo
me
try
En
ric
hm
en
t8-7
Do
t P
ro
du
ct
Th
e d
ot
pro
du
ct o
f tw
o v
ect
ors
rep
rese
nts
how
mu
ch t
he v
ect
ors
poin
t in
th
e d
irect
ion
of
each
oth
er.
If
� v is
a v
ect
or
rep
rese
nte
d b
y ⟨a
, b
⟩ an
d � u
is a
vect
or
rep
rese
nte
d b
y ⟨c, d
⟩, t
he
form
ula
to f
ind
th
e d
ot
pro
du
ct i
s:
� v ·
� u = ac +
bd
Look
at
the f
oll
ow
ing e
xam
ple
:
Gra
ph
th
e v
ect
ors
an
d f
ind
th
e d
ot
pro
du
ct o
f � v
an
d � u
if
� v =
⟨3,
-1
⟩ an
d � u
= ⟨
2,
5⟩.
� v ·
� u =
(3)(
2)
+ (
-1)(
5)
or
1
Gra
ph
th
e v
ecto
rs a
nd
fin
d t
he
do
t p
ro
du
cts
.
1
. � v
= ⟨
2,
1⟩
an
d � u
= ⟨
-4,
2⟩
2. � v
= ⟨
3,
-2
⟩ an
d � u
= ⟨
1,
4⟩
Th
e d
ot
pro
du
ct i
s
T
he d
ot
pro
du
ct i
s
3. � v
= ⟨
0,
3⟩
an
d � u
= ⟨
2,
4⟩
4. � v
= ⟨
-1,
4⟩
an
d � u
= ⟨
-4,
2⟩
Th
e d
ot
pro
du
ct i
s
T
he d
ot
pro
du
ct i
s
5. N
oti
ce t
he a
ngle
form
ed
by t
he t
wo v
ect
ors
an
d t
he c
orr
esp
on
din
g d
ot
pro
du
ct.
Is t
here
an
y r
ela
tion
ship
betw
een
th
e t
yp
e o
f an
gle
betw
een
th
e t
wo v
ect
ors
an
d t
he s
ign
of
the
dot
pro
du
ct?
Mak
e a
con
ject
ure
.
y
xO
y
xO
y
xO
y
xO
y
xO
Yes,
there
is a
rela
tio
nsh
ip b
etw
een
th
e t
yp
e o
f an
gle
betw
een
th
e t
wo
vecto
rs a
nd
th
e s
ign
of
the d
ot
pro
du
ct.
Wh
en
th
e a
ng
le f
orm
ed
is a
cu
te,
the s
ign
of
the d
ot
pro
du
ct
is p
osit
ive a
nd
wh
en
th
e a
ng
le f
orm
ed
is
ob
tuse,
the s
ign
of
the d
ot
pro
du
ct
is n
eg
ati
ve.
-5
-6
12
12
Answers (Lesson 8-7)
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.
An
sw
ers
Chapter 8 A25 Glencoe Geometry
Chapter 8 Assessment Answer KeyQuiz 1 (Lessons 8-1 and 8-2) Quiz 3 (Lessons 8-5 and 8-6) Mid-Chapter Test
Page 51 Page 52 Page 53
Quiz 2 (Lessons 8-3 and 8-4)
Page 51
Quiz 4 (Lesson 8-7)
Page 52
Part I
Part II
1.
2.
3.
4.
5.
8 √ " 3
x = √ "" 85 ,
y = 2 √ "" 15
x = √ "" 17 , y = 81
− 8
√ "" 137
acute
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
3 √ " 2
2 √ " 3
57.8
14.6
20 √ " 3 + 20 in.
0.7880
27
3.3 ft
27.1
22.0
1.
2.
3.
4.
5.
∠QPR
3.9
m∠B = 104,m∠C = 27,b = 23.2
m∠R = 88,m∠T = 53,m∠S = 39
52°
1.
2.
3.
4.
5.
24.8 units, 220°
31.4 units, 158°
6 mph, south
c"d"
c" d"+
k" k"
-m"
-m"
6.
7.
8.
9.
10.
x = 9, y = 3 √ " 3
x = 10 √ " 3 ,
y = 10 √ " 6
x = 12 √ " 2 ,
y = 24 √ " 2
right
54
1.
2.
3.
4.
5.
A
H
B
H
D
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ies, In
c.
Chapter 8 A26 Glencoe Geometry
Chapter 8 Assessment Answer KeyVocabulary Test Form 1
Page 54 Page 55 Page 56
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
geometric mean
Pythagorean triple
false,trigonometric
ratio
true
angle of depression
trigonometry
sine
tangent
A method to describe a vector using its
horizontal and vertical change from its inital
to terminal point.
In a right triangle, the sum of the squares of
the measures of the
legs equals the square
of the measure of the
hypotenuse.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
D
F
A
G
D
G
C
G
C
H
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
B:
C
F
A
G
D
F
B
F
C
G
69
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.
Answers
Chapter 8 A27 Glencoe Geometry
Chapter 8 Assessment Answer KeyForm 2A Form 2B
Page 57 Page 58 Page 59 Page 60
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
D
J
A
H
D
J
A
G
B
H
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
B:
B
F
C
G
D
H
C
G
D
H
56.4 ft
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
A
J
D
G
D
F
A
H
B
H
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
B:
B
H
C
G
D
G
A
G
B
J
73.2 ft
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Chapter 8 A28 Glencoe Geometry
Chapter 8 Assessment Answer KeyForm 2C
Page 61 Page 62
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
√ "" 10
13
2 √ " 7
√ """ 4000 or 20 √ "" 10
√ "" 300 or 10 √ " 3
11 √ " 2
2 √ " 3
x = 6, y = 12
9.7
69
68
12.
13.
14.
15.
16.
17.
18.
19.
20.
B:
267.9m
11°
43.2
14.3
15.6 ft
20
33.0
√ "" 170 ≈ 13 units; 237.5°
1
s
t
s t+
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Answers
Chapter 8 A29 Glencoe Geometry
Chapter 8 Assessment Answer KeyForm 2D
Page 63 Page 64
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
3 √ "" 10
26
√ "" 39
√ """ 7300 or 10 √ "" 73
√ "" 432 or 12 √ " 3
15 √ " 2
7 √ " 3
x = 12, y = 24
8.6
70°
67°
352.7 m
9°
32.3
6.2°
15.2 ft
19
38.2
√ "" 109 ≈ 10.4 units, 253.3°
6 or 10
d
-f
d f-
12.
13.
14.
15.
16.
17.
18.
19.
20.
B:
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Chapter 8 A30 Glencoe Geometry
Chapter 8 Assessment Answer KeyForm 3
Page 65 Page 66
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
√ " 6
− 9
2
3
2 √ "" 10
12 −
5
12.5
250 km
right
12 √ " 6
24 + 8 √ " 3
x = 5 √ " 3 ,
y = 10 √ " 3
(-4 + 8 √ " 3 , -2)
or (-4 - 8 √ " 3 , -2)
13.
14.
15.
16.
17.
18.
19.
20.
B:
x = 5 √ " 6
− 3 , CD = 5
√ " 6 −
3
+ 5 √ " 2 + 12
147 ft
2 ft
37.4
253.7 yd
37° or 143°
31.4 units, 261°
71.1 ft and 84.5 ft
a
ab
-b
a b-
a b+
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Answers
Chapter 8 A31 Glencoe Geometry
Chapter 8 Assessment Answer KeyExtended-Response Test, Page 67
Scoring Rubric
Score General Description Specific Criteria
4 Superior
A correct solution that
is supported by well-
developed, accurate
explanations
• Shows thorough understanding of the concepts of geometric mean,
special right triangles, altitude to the hypotenuse theorems,
Pythagorean Theorem, solving triangles, SOH CAH TOA, Law of
Sines, and Law of Cosines.
• Uses appropriate strategies to solve problems.
• Computations are correct.
• Written explanations are exemplary.
• Figures are accurate and appropriate.
• Goes beyond requirements of some or all problems.
3 Satisfactory
A generally correct solution,
but may contain minor flaws
in reasoning or computation
• Shows an understanding of the concepts of geometric mean,
special right triangles, altitude to the hypotenuse theorems,
Pythagorean Theorem, solving triangles, SOH CAH TOA, Law of
Sines, and Law of Cosines.
• Uses appropriate strategies to solve problems.
• Computations are mostly correct.
• Written explanations are effective.
• Figures are mostly accurate and appropriate.
• Satisfies all requirements of problems.
2 Nearly Satisfactory
A partially correct
interpretation and/or
solution to the problem
• Shows an understanding of most of the concepts of geometric
mean, special right triangles, altitude to the hypotenuse theorems,
Pythagorean Theorem, solving triangles, SOH CAH TOA, Law of
Sines and Law of Cosines.
• May not use appropriate strategies to solve problems.
• Computations are mostly correct.
• Written explanations are satisfactory.
• Figures are mostly accurate.
• Satisfies the requirements of most of the problems.
1 Nearly Unsatisfactory
A correct solution with no
supporting evidence or
explanation
• Final computation is correct.
• No written explanations or work is shown to substantiate the final
computation.
• Figures may be accurate but lack detail or explanation.
• Satisfies minimal requirements of some of the problems.
0 Unsatisfactory
An incorrect solution
indicating no mathematical
understanding of the
concept or task, or no
solution is given
• Shows little or no understanding of most of the concepts of
geometric mean, special right triangles, altitude to the hypotenuse
theorems, Pythagorean Theorem, solving triangles, SOH CAH
TOA, Law of Sines and Law of Cosines.
• Does not use appropriate strategies to solve problems.
• Computations are incorrect.
• Written explanations are unsatisfactory.
• Figures are inaccurate or inappropriate.
• Does not satisfy requirements of problems.
• No answer given.
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Chapter 8 A32 Glencoe Geometry
Chapter 8 Assessment Answer KeyExtended-Response Test, Page 67
Sample Answers
1. 10
− 6 =
6 −
x
10x = 36
x = 18
− 5
2a. No, his work is not correct. The triangle is not a right triangle so he cannot use the altitude to the hypotenuse theorem. Instead he must use the 30°-60°-90° triangle relationships and obtain
x = 2 √ $ 3 .
b. Using the 30°-60°-90° triangle relationships, RQ = 4, and therefore PQ would have to equal 8, so PS = 8 - 2 or 6.
c. No, the sides are not in a ratio of 1:1: √ $ 2 .
3. If you were finding the length of a side x you would use sin of the given angle. If you are finding an angle x you would use the s in -1 of the ratio of two sides to find the angle. For example, if
sin x = 3 −
4 , use s in -1
3 −
4 to find x. If
sin 20° = x −
5 find the sine of 20° and
multiply by 5 to find x, the length of the unknown side.
4. angle of depression
angle of elevation
Student should draw a right triangle and label the angle of elevation up from the horizontal and the angle of depression down from the horizontal. The angle of elevation has the same measure as the angle of depression.
5.
150A C
B
20°
110°ca
Let m∠B = 110, m∠A = 20, and b = 150. Use the Law of Sines to find the missing lengths. The length of a is found using
sin 110°
− 150
= sin 20°
− a . The third angle is
found by evaluating 180 - (m∠B + m∠A). In this problem, m∠C = 50. The length of
c is found by using sin 110°
− 150
= sin 50°
− c .
In this problem, c ≈ 122.3.
6. No, her plan is not a good one. Irina should use the Law of Cosines to find angle B. Then she can use the Law of Sines to find either angle A or angle C and subtract the two angles she finds from 180 degrees to get the measure of the third angle.
In addition to the scoring rubric found on page A31, the following sample answers may be used as guidance in evaluating open-ended assessment items.
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Answers
Chapter 8 A33 Glencoe Geometry
Chapter 8 Assessment Answer KeyStandardized Test Practice
Page 68 Page 69
1. A B C D
2. F G H J
3. A B C D
4. F G H J
5. A B C D
6. F G H J
7. A B C D
8. F G H J
9. A B C D
15.
16.
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
1 8 0.
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
9
8
7
6
5
4
3
2
1
0
1 4 5.
10. F G H J
11. A B C D
12. F G H J
13. A B C D
14. F G H J
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Chapter 8 A34 Glencoe Geometry
Chapter 8 Assessment Answer KeyStandardized Practice Test
Page 70
17.
18.
19.
20.
21.
22.
23a.
23b.
23c.
Def. of segments
HJ + JK = HK;KL + LM = KM
m∠1 = 32, m∠2 = 82,
m∠3 = 66, m∠4 =
114, m∠5 = 57
7
19
∠3, ∠4, ∠5, ∠7,∠ABC
yes, both are 30°- 60°- 90°
triangles; AASimilarity
36
2:1