basic two dimensional geometry with … two...theorem 1 (aa) if two angles of a triangle are...
TRANSCRIPT
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BASIC TWO DIMENSIONAL GEOMETRY
WITH APPLICATIONS
LUONG LE
GEOMETRY
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This book is dedicated to my counselor and a dear friend,
Julie Visner.
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CHAPTER 5
5.1 Similar Triangles
5.2 The Pythagorean Theorem
5.3 Special Right Triangle
5.4 Laws of sine and cosine
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Section 5.1 Similar Triangles. Outline: Definition.
Prove two triangles are similar.
Proportional segments.
Applications.
1) Definition (symbol ∾ )
Two triangles are similar if corresponding angles are congruent, denoted CASTC, and corresponding sides are proportional, denoted CSSTP.
Example 1
41 ∠≅∠ ; 52 ∠≅∠ ; 63 ∠≅∠ and xr
hH=
In general: Two figures are similar if they have the same shape, but they’re different in the size.
Example 2
5 4
6 3
1 2
x H
h
r
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2) Prove two triangles are similar.
Theorem 1 (AA)
If two angles of a triangle are congruent to two angles of another triangle, then two triangles are similar.
(See the proof in the example 6 of section 5.4)
Example 3
Given: ΔABC and ΔEFC with
∠ B and ∠ F are rt. s' ∠
Prove: ΔABC ~ ΔEFC
STATEMENTS REASONS
1) ∠ B and ∠ F are rt. s' ∠ 1) Given
2) FB ∠≅∠ 2) All rt. ∠ ’s are ≅
3) CC ∠≅∠ 3) Identity
4) ΔABC ∾ ΔEFC 4) AA
Joke: How do you prove speeches of two first ladies Obama and Trump similar?
Two speech writers: AA
A
B F
E
C
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3) Proportional segments
a) Theorem 2
If a line intersects two sides of a triangle and it’s parallel to the third side, then the line divides two sides of triangle proportionally.
Given: DC||EB
Prove: EDAE
BCAB
=
STATEMENTS REASONS
1) DC||EB 1) Given
2) 21 ∠≅∠ 2) Cor. ∠ ’s are ≅
3) AA ∠≅∠ 3) Identity
4) ADC∆ ∾ ΔAEB 4) AA
5) ADAE
ACAB
= 5) CSSTP
6) AEAD
AEABAC
AB−
=−
6) Subtraction property of the proportion
7) EDAE
BCAB
= 7) Segment addition property
1 E
D C
B
A
2
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b) Theorem 3 (Converse):
If a line divides two sides of triangle proportionally, then it’s parallel to the third side.
Given: EDAE
BCAB
= Prove: DC||EB
Proof by contradiction (contra position)
STATEMENTS REASONS
1) EDAE
BCAB
= (*) 1) Given
2) DC||EB , 2) Assumption
3) Draw DC||EF and AF ≠ AB 3) Construction
4) EDAE
FCAF
= (**) 4) If a line is parallel 3rdside of Δ , it divides two other sides of Δ proportionally (Theorem 2)
5) AEED
AEABBC
AB+
=+
;
EDAEAE
FCAFAF
+=
+
5) Addition property of proportion for (*) and (**)
6) ADAE
ACAF
= ; ADAE
ACAB
= 6)Segment addition postulate
7) ACAB
ACAF
= 7) Transition
8) AF = AB, a contradiction 8) Properties of proportion
9) DC||EB 9) By contraposition proof
c) Theorem 4 (SAS ∾) If two triangles have one pair of included congruent angles and two pairs of proportional sides, they are similar. (See homework exercise)
d) Theorem 5 {SSS ∾) If two triangles have three pairs of proportional they are similar. (See homework exercise.)
E
D C
B
A
F
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e) Angle bisector Theorem
The angle bisector divides the opposite side into two segments proportionally to the two sides of the bisected angle.
Given: PQR∆ and 21 ∠≅∠
Prove: SQRS
PQRP
=
STATEMENTS REASONS
1) 21 ∠≅∠ 1) Given
2) Draw PS||TQ 2)Construction
3) SQRS
PTRP
= 3) If a line is parallel 3rdside of Δ , it divides two other sides of Δ proportionally (Theorem 2)
4) 24 ∠≅∠ 4) Cor. ∠ ’s are ≅
5) 31 ∠≅∠ 5) Alt. int. ∠ ’s are ≅
6) 43 ∠≅∠ 6)Transition
7) PQ = TP 7) In a Δ , opp. sides of ≅ ∠ ’s are≅
8) SQRS
PQRP
= 8) Substitution 7) into 3)
Example 3: PT bisects P∠ . Find x
Solution:
By the angle bisector theorem, we have
x10x
57
−=
X = in83.51270
≈
P
R Q S
T
1 2
3
4
P
T R Q
7 in 5 in
x
10 in
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f) Ceva’s theorem From a random interior point D of a triangle ABC, we draw three segments connecting D to the vertices A, B, and C. The three segments intersect the three sides of the triangle ABC at E, F, and G. Then
1GACG
ECBE
FBAF
=⋅⋅
The proof is in the Appendix e) Corollary If two transversals are cut by three or more parallel lines, then two transversals are divided proportionally by the parallel lines.
Given: AD‖BE‖CF
Prove: ABBC
= DEEF
STATEMENTS REASONS
1) AD‖BE‖CF 1) Given
2) Draw AF 2)Construction
3) ABBC
= AGGF
; GFAG
= EFDE
3) If a line is parallel 3rdside of Δ , it divides two other sides of Δ proportionally (Theorem 2)
4) AGGF
= DEEF
4) Property of proportion
5) ABBC
= DEEF
5) Transition
A
B C
E
G
F
D
C
A
B
D
E
F
G
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4) Applications
Example 4: How high should Joe build the fence to block the view from the top of his neighbor’s window to the top of his own one?
ΔABC ~ ΔEFC . Why?
⇒=FCBC
EFAB
⇒=38
x1.8 m675.0
8(1.8)(3)
==x
Thus, Joe has to build the fence that is higher than 2.675 m
A
B F
E
C
x
8m
3 m
1.8m
https://www.google.com/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&cad=rja&uact=8&ved=0ahUKEwiK2uPOtffRAhVG0mMKHRvUCHQQjRwIBw&url=http://www.helpingwithmath.com/by_subject/geometry/geo_similar_triangles_8g4.htm&bvm=bv.146094739,d.cGc&psig=AFQjCNGuk7w-zuTXgrNDi3zQqdA1oNbbmw&ust=1486330904481419
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Example 5: Fundamental lens equation: Im1
Obj1
f1
+=
The focal length f is the distance from the focal point to the lens. The height of the object is a; the height of the image is b; the distance from the object to the lens is Obj; the length form the lens to the image is Im. We have two pairs of similar triangles: ΔABC ~ ΔODC and ΔHOG ~ ΔFEG
Solution:
ΔABC ~ ΔODC ; ΔHOG ~ ΔFEG
COAC
ODAB
= ; GFOG
EFOH
= (CSSTP)
Let AB = HO = a; DO = EF = b; CO = OG = f; BH = Obj, OF = Im Then AC = Obj – f; GF = Im – f. Substituting, we get
ffObj
ba −= ;
f Imf
ba
−=
Transition
ffObj
fImf −
=−
Cross Multiply f2 = (Im – f )(Obj – f) Distribution
⇒+−−= 22 f(Obj)(f)(Im)(f)(Im)(Obj)f (Im)(Obj) = (Im)(f) + (f)(Obj) (*) Now, divide (*)by (f) (Im)(Obj), we obtain
Im1
Obj1
f1
+=
b
Obj
Im
f f
H
F
D
A
B
C
E
G a
O
lens
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Example 6: The minimum distance
A firefighting helicopter is at a location 45 miles downstream from a fire. The helicopter is 20 miles away from a big river and the fire is 10 miles away from the river. The pilot has to fill the helicopter with the water from the river and she drops it on the fire. To what point on the river, should she fly the copter to minimize the distance
Solution:
The minimum distance occurs when the segments D1 and D2 form a straight line
BDCHAB ∠≅∠ ; DBCHBA ∠≅∠ ⇒ ΔABH ~ ΔDBC
1020
x x- 45
= (CSSTP)
Cross Multiply
10(45 – x) = 20 x⇒ x = 15 miles.
A
H
x 45 - x
10 20
D1
D2
10 D2
45
B
C
D
F
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HOMEWORK 5.1
1) Does the similar relation have the reflexive property? The symmetric property? The transitive property?
2) What is CASTC?
3) What is CSSTP?
4) Given: ΔABC ~ ΔEFG ; 054Am =∠ 062Cm =∠
AB = 21cm, DE =14cm, DF = 12cm, BC= 15 cm
Find: a) Em∠ b) AC c) EF
5) Given: ΔABC ~ ΔADE ; 036Bm =∠
AD = 15cm, AC =25cm, EB = 20cm.
Find: a) Dm∠ b) AE
6) Given: ΔPQR ~ ΔSTR ; 044Sm =∠
RS = 6cm, PR =27cm, RT = 2.4cm
Find: a) PRQm∠ b) RQ
7) Fundamental lens equation: Im1
Obj1
f1
+= . The focal length f is the distance from the focal point to
the lens. The height of the object is a; the height of the image is b; the distance from the object to the lens is Obj; the length form the lens to the Image is Im.
A wide range 28mm lens on 35 mm camera has a focal length f = 50mm (about 2 inches). How far from the film will the lens be located if they are focus on an object 3m away?
A
D
F E
C B
A
D
C
B E S
R
Q P
T
28mm lens
O b
Obj
Im
f f
H
F
D
A
B (object)
C E(image)
G a
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8) A ball is served from the center of the baseline into the deuce court. The ball is hit 8ft above the ground and the net is 3ft high. If the ball just clears the top of the net, how far from the base of the net will the ball land?
9) The top of the tree reflects from a mirror so that DBEABC ∠≅∠ . Find the height of the tree in the figure if AB = 24ft, BD = 2ft 6 in and DE = 5ft 9 in
10) A 5ft young girl who walks away from a pyramid casts a shadow of 9ft long. If she is at the distance of 859ft from the pyramid at that moment, how high is the pyramid?
11) The garage ceiling is of 10ft above the ground and the opener is mounted 10ft away from the garage door. If the light casts 6ft beyond the garage door, how far is the garage door above the ground?
78ft
8ft
Deuce court 3ft
A D
B
E
A
C
9ft
5ft
859ft
10ft
6ft
10ft
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12) A firefighting helicopter is at a location 45 miles downstream from a fire. The helicopter is 20 miles away from a big river and the fire is 10 miles away from the river. The pilot has to fill the helicopter with the water from the river and she drops it on the fire. To what point on the river, should she fly the copter to minimize the distance.
13) The segment PT bisects P∠ . If PQ = 5in, PR = 7 in, and QR = 10 in, find TR
14) The segment PT bisects P∠ . If PQ = 3cm, PR = 5cm, QT = 2cm. Find TR.
15) The segment PT bisects P∠ . If PQ = x + 6 cm, PR = 2x +1 cm, QT = 3cm, and TR = x – 2 cm, find x.
16) The segment PT bisects P∠ . If PR = 3x – 1 in, PQ = 2x in, QT = x in, and TR = x + 1in, find x.
17) The rhombus BDEF is inscribed in a triangle ABC. If AE = 10 mm, CE = 6mm. Find x.
A
H
x 45 - x
10 20
D1
D2
10 D2
45
B
C
D
F
C
D
B
A
E
F
x
P
T R Q
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18) A square ABFG with sides of length 2 in. rests on a square BCDE with sides of length 6 in. Find the perimeter of trapezoid ABCD.
Exercise 18 Exercise 19
19) Given: DE||AB
Prove: ADC∆ ~ ADC∆
STATEMENTS REASONS
1) DE||AB 1)?
2) 21 ∠≅∠ and 43 ∠≅∠ 2)?
3) ADC∆ ~ ADC∆ 3)?
20) Given: NM ∠≅∠
Prove: MQR∆ ~ NPR∆
STATEMENTS REASONS
1) NM ∠≅∠ 1)?
2) 21 ∠≅∠ 2)?
3) MQR∆ ~ NPR∆ 3)?
A
D E
C
B 1
2
3
4
M
R
Q
P
N
1 2
B E
D C
A
F
G 2in
6in
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21) If two triangles have one pair of included congruent angles and two pairs of proportional sides, they are similar.
Given: DCBC
CEAC
= ; 21 ∠≅∠
Prove: ACB∆ ~ ECD∆
STATEMENTS REASONS
1) DCBC
CEAC
= 1)?
2) DE||AB 2)?
3) 43 ∠≅∠ 3)?
4) 21 ∠≅∠ 4)?
3)? 5)?
22) Given: DE||CB
Prove: ACAEABAD ⋅=⋅
23) Given: MNPQ is a parallelogram
QPMS ⊥ and QPMS ⊥
Prove: MQS∆ ~ MTN∆
A
D E
C
B
1
2
3
4
A
D
C
B E
P
M N
Q S
T
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24) If two triangles have three pairs of proportional they are similar.
Given: EFBC
DEAB
DFAC
==
Prove: ABC∆ ~ DEF∆
25) Given: PQRS is a parallelogram.
Prove: TSR∆ ~ RQU∆
STATEMENTS REASONS
1) EFBC
DEAB
DFAC
== 1)?
2) DE||AB ; DF||AC and FE||CB 2)?
3) F1 ∠≅∠ and C1 ∠≅∠ 3)?
4) FC ∠≅∠ 4)?
5) EB ∠≅∠ 5)Repeat step 3) and step 4)
6) ABC∆ ~ DEF∆ 6)
P
S R
Q
T
U
1
A
D
F E
C B
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26) Given: DB ∠≅∠
Prove: ABC∆ ~ EDC∆
27) Given: DF||AB and FG||AD
Prove: ABC∆ ~ EFG∆
28) Given: DF||AB and AF||CE
Prove: AGDECEAB ⋅=⋅
29) The distance across a pond is to be measured indirectly by using similar triangles. If XY = 160ft, YW = 40ft, TY = 120ft, and WZ = 50f, find XT.
30) Using the measurements given to find the distance across the river.
Exercise 29 Exercise 30
A
D
C B E
A
B C
D
E
F
G
B
A C D
E
F
G
X
T
W
Z
Y POND
C
F
RIVER
A
D
B
30ft
25ft
4ft
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31) Prove that the altitude drawn to the hypotenuse of a right triangle separates the right triangle into two right smaller triangles that are similar to each other and the original right triangle.
32) In triangle ABC with the interior point D, if CG = 5in, GB = 10in, BF = 9in, HT = 3 in, and CE = 4in, find AE?
34) The truncated conical solar collector is aimed directly at the sun as shown in the figure. If R = 12mm, H = 15.39mm, h = 5mm, find r.
SUMMARY
35) What is the definition of similar triangles?
36) How to prove two triangles are similar?
37) State the angle bisector theorem.
38) State Theorem 2 (AA)
B
A
C D
R
r
H
h
A
B C
E
G
F
D
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5.2 Pythagorean Theorem Outline: Theorems of the right similar triangles
Pythagorean Theorem
Applications
1) Theorems of the right similar triangles
a) Theorem 1
The altitude from a right angle divides the right triangle into two smaller right triangles that are similar to each other and to the original triangle.
Given: PQR∆ with rt. P∠ ; RQPT ⊥
Prove: PQR∆ ~ TPQ∆
PQR∆ ~ TPR∆
TPQ∆ ~ TPR∆
STATEMENTS REASONS
1) P∠ is rt. ∠ ; RQPT ⊥ 1) Given
2) 2 and 1 ∠∠ are rt. ∠ 2) ⊥ lines form rt. s'∠
3) 2P ∠≅∠ ; 1P ∠≅∠ 3) All rt. ∠ ’s are ≅
4) QQ ∠≅∠ ; RR ∠≅∠ 4) Identity.
5) PQR∆ ~ TPQ∆ ; PQR∆ ~ TPR∆ 5) AA
6) TPQ∆ TPR∆ 6)Transition
R
P
T Q 1 2
-
b) Theorem 2
The length of the altitude to the hypotenuse of a right triangle is the geometric mean of the lengths of the segments of the hypotenuse.
Given: PQR∆ with rt. P∠ ; RQPT ⊥
Prove: QTPT
PTRT
=
STATEMENTS REASONS
1) PQR∆ with rt. P∠ ; RQPT ⊥ 1) Given
2) TPQ∆ ~ TPR∆ 2)Theorem1
3) QTPT
PTRT
= 3) CSSTP
c) Lemma
The length of each leg of a right triangle is the geometric mean of the lengths of the hypotenuse and the length of the segments of the hypotenuse adjacent to the legs.
RTPR
PRRQ
= or QTPQ
PQRQ
=
The proof of the lemma is left for exercise
R
P
T Q 1 2
R
P
T Q 1 2
-
2) Pythagorean Theorem
The square of the length of the hypotenuse equals to the sum of the squares of the lengths of the legs.
Given: ABC∆ with rt. A∠ ; BC = c; AB = b; AC = a
Prove: 222 cba =+
STATEMENTS REASONS
1) ABC∆ with rt. A∠ ; BC = c; AB = b; AC = a 1) Given
2) Draw CBAD ⊥ ; CD = x; DB = y ; x + y = c 2) Construction
3) xa
ac= ;
yb
bc=
3) Lemma 3
4) cxa =2 cyb =2
4)Cross multiply
5) )(22 yxcba +=+ 5) Add and factor
6) 222 cba =+ 6) Substitution
C
A
D B x y
a b
c
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3) Applications
Example 1: A drawbridge that is 104 ft in length is raised at its midpoint so that the upmost points are 8ft apart. How far has each of midsection been raised?
22 4852 −=x = 20
Example 2: A hangar has the shape of an isosceles trapezoid surmounted by an isosceles triangle.
If AE = 30 ft, BD = 24 ft, CM= 5 ft, BH = 8 ft. Find EF, DM, CB, and AB.
This problem is left for exercise.
Joke: Alternative fact
In the Black History month, Frederick Douglass, a PhD candidate student of the Trump University under the mentors of Dr. Pepper and Kellyanne Conway, wrote the dissertation on the alternative facts proving that
(Power of the Congress)2+ (Power of the Judiciary)2 = (Power of the Presidency)2
A
B
C
D
E
F H
M
52
48
x
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HOMEWORK 5.2
1) Find the length of BC if
a. AB = 8”and AC = 6”
b. AB = 11”and AC = 9”
2) Find the length of AC if
a. AB = 8 cm and BC = 14cm
b. AB = 11”and BC = 20”
3) Find the length of AB if
a. AC = 6”and BC = 14”
b. AC = 9mm and BC = 13mm
4) Find the length of AH if
a. BC = 13”, AB = 12”, and AC = 5”
b. BC = 10cm, AB = 8 cm, and AC = 6 cm
5) A rectangle has a width of 12cm and the length of the diagonal is 34cm. Find the length of the rectangle.
6) A rectangle has a width of x, a length of x + 1 and a length of the diagonal is 2x + 3. Find the length of the rectangle.
7) A right triangle has the legs of x + 1 and x + 3and a hypotenuse 2x. Find x.
8) The lengths of diagonals of a rhombus are 6ft and 8ft. Find the length of its sides.
9) A kite has a side of12 cm and another side of 16 cm. If the length of one diagonal is 6 cm, how long is the other diagonal?
10) A rhombus has a side of12 cm. If the length of one diagonal is 18 cm, how long is the other diagonal?
11) An isosceles right triangle has a hypotenuse of 14”. How long is each leg?
12) If a leg of an isosceles right triangle is 4ft, what is the length of its hypotenuse?
13) In a right triangle ABC with the right A∠ , if M is the midpoint of AB , the hypotenuse BC = 10’, and AC = 8’, find CM.
14) In a right triangle ABC with the right A∠ , if M is the midpoint of AB , the hypotenuse BC = 13m, and AC = 5m, find CM.
A
B C H
-
15) In a right triangle ABC with the right A∠ , if M and N are the midpoints of AB and BCrespectively, AB = 14m, and BC = 6m, find MN.
16) In a right triangle ABC with the right A∠ , if M and N are the midpoints of AB and BCrespectively, AB = 8m, and BC = 17m, find MN.
17) 5ft fence is 4ft away from a house. A 10 foot ladder is against the house from the ground and across the top of the fence. How far from the fence will the ladder be?
18) A piece of a rectangular paper can be fold to create 5” x 7” envelope as shown in the figure. Find the dimension of the rectangular paper.
19) Using the measurements as shown in the figure to find the height of the pyramid?
20) The building is 30ft tall and casts a shadow 8ft wide in front of the building. If Mindy is 5’ 6” tall, how far away from the building can she stand and still be able in the shade?
30’
8’
-
21) A hot air balloon that is 21ft above the ground is held by the crew 32ft from the point directly beneath the basket (see the figure.} How long is the rope?
22) A wind holds a kite 30ft above the ground in a place 40ft across the ground (see figure.) What is the length of the string?
23) A small boat that is held by 40ft rope is 18ft below the level of a pier. How far is from the pier across the water to the boat.
24) A hangar has the shape of an isosceles trapezoid surmounted by an isosceles triangle.
If AE = 30 ft, BD = 24 ft, CM= 5 ft, BH = 8 ft. Find EF, DM, CB, and AB.
25) If AC = 6, BC = 13, and AB = 15, find HC?
26) If AC = 6, BC = 13, and AB = 15, find HC?
27) A square EFGH has its vertices on the sides of the square ABCD. Show that ΔCHGΔBGF ≅
28) A square EFGH has its vertices on the sides of the square ABCD. If BG = 5m, GC = 12m. Find FG
Exercise 25 and 26 Exercise 27 and 28
21’ 32’
30’
40’
40’ 18’
G
F
E D
C B
A
H
C
A B H
A
B
C
D
E
F H
M
-
29) Prove that the altitude from a right angle divides the right triangle into two smaller right triangles that are similar to each other and to the original triangle.
30) Prove that the length of the altitude to the hypotenuse of a right triangle is the geometric mean of the lengths of the segments of the hypotenuse.
31) Prove Pythagorean Theorem.
32) Prove that the length of each leg of a right triangle is the geometric mean of the lengths of the hypotenuse and the length of the segments of the hypotenuse adjacent to the legs.
Given: PQR∆ with rt. P∠ ; RQPT ⊥
Prove: RTPR
PRRQ
=
STATEMENTS REASONS
1) PQR∆ with rt. P∠ ; RQPT ⊥ 1) Given
2) TRP∆ ~ PRQ∆ 2)?
3) RQPR
PRRT
= 3)?
4) RTPR
PRRQ
= 4)?
SUMMARY: Complete each sentence
33) The _________from a right angle divides the right triangle into two smaller right triangles that are __________ to each other and to the ________triangle.
34) Prove that the length of the ___________ to the hypotenuse of a right triangle is the _________ mean of the lengths of the __________ of the hypotenuse.
35) The length of each ______of a right triangle is the geometric mean of the lengths of the __________ and the length of the __________ of the hypotenuse ________ to the legs.
36) State the Pythagorean Theorem.
R
P
T Q
1 2
-
5.3 Special Right Triangles
Outline: The 45-45-90 triangle
The 30-60-90 triangle 1) The 45-45-90 triangle
In the 45-45 -90 triangle, the legs are opposite to the congruent angles. Thus, they are congruent. By
the Pythagorean Theorem, we have the hypotenuse has the length equal to the product of 2 and the length of the leg.
2) The 30-60-90 triangle
In the 30-60 -90 triangle, the hypotenuse is double the length of the shortest leg. The longer leg has the
length equal to the product of 3 and the length of the shortest leg.
Example 1
An 8ft ram that is raised 4 ft to the door of a truck is used to unload the furniture. What is the slope of the ram?
The problem is left as exercise.
a
a
450
450
a 2
600
300
-
Example 2:
The segments QS and QT trisect the rt.∠ A. Find RT, SP, and PR.
Solution.
PQ = RQ 3 ⇒ cm3
8RQ =
PR = (2) RQ⇒ cm3
16PR =
Since QS and QT trisect ∠ A, ΔRTQ and TSQ∆ are 30 – 60- 90 s'∆
cm3
42
RQRT ==
Since m∠ SQP = 300, ΔPSQ is isosceles.
cm3
8RQQSSP ===
R
Q P
S
T
30o
8
-
Homework 5.3
For 450 -450 right triangle.
1) If AC = 2cm, find AB.
2) If AB = 5”, find BC.
3) If AB = a units, find AC.
4) If BC = b units, find AB. Exercise 5 – 14
For 300 -600 right triangle.
5) If AC = 5cm, find AB and BC
6) If AB = 20”, find BC and AC
7) If BC = 10’, find AB and AC
8) If AB = a units, find AC and BC
9) If BC = b units, find AB and AC.
10) If AC = c units, find BC and AB.
11) If AB = 33 m, find AC and BC Exercise 5 – 13
12) If BC = 12ft, find AB and AC.
13) If AC = 4cm, find BC and AB.
Using the measurements as shown to find
15) RQ and PQ.
16) AB, AC and BD. Exercise 15
Exercise 16
C
B
A 450
A B
C
300
600
Q
P
300 450
S R
12cm
300
450
A
4” D
C B
-
17) The truncated conical solar collector is aimed directly at the sun as shown in the figure. If060ECDm =∠ , AB = 12”, find h.
18) Given: OT bisects RQP∠ , 030Pm =∠ and PQ = 12 cm
Prove: PT and TR.
19) Given: OT bisects RQP∠ , RQ = 6m and PQ = 12 m
Prove: PT and TR.
20) Given: OT bisects RQP∠ , RQ = 4mm and PR = 34 mm Exercise 18 – 22
Prove: PT and TR.
21) Given: OT bisects RQP∠ , 030Pm =∠ PR = 32 in
Prove: PT and TR.
22) Given: OT bisects RQP∠ , 030Pm =∠ RQ = 5ft
Prove: PT and TR.
23) Given: PQ bisects MPN∠ , QN bisects MNP∠ , MNP∆ is equilateral, QN = 4”
Prove: MN.
24) Given: MNP∆ is an isosceles right triangle, BC = 220 cm
M and N are mid points of AB and BC
Prove: MN.
25) An 8ft ram that is raised 4ft to the door of a truck is used to unload the furniture. What is the slope of the ram.
r
h R
12”
B
A
600
D
C E 300
P
Q R
T
M
N
Q
P
B
A C
M N
-
26) The segments QS and QT trisect the rt.∠Q. Find RT, SP, and PR.
27) To get the distress swimmer, the lifeguard must run some distance along the beach at the rate of 5m/s, enter the water at the rate of 1.7m/s. If d = 380m, c = 76m, and Ɵ = 600, find the distance of running and the distance of swimming.
28) An island is 4mi offshore in a large bay. A water pipeline is to be run from a water tank on the shore to the island with angle Ɵ = 300, as shown in the figure. The pipeline costs $40,000 per mile in the ocean and $20,000 per mile on the land. Use the measurements as shown in the figure to find the total cost of the pipelines.
Exercise 27 Exercise 28
29) Given: Isosceles trapezoid ABCD with DC = 6ft, 0120Am =∠ DE bisects ADC∠ , CE bisects EBC∠ Find: The perimeter of ABCD
30) Given: MN = NP = PQ = 6”, 0120PmNm =∠=∠ Find: MP and MQ.
31) Given: ABCE is a square, CDE∆ is an isosceles triangle, AB = 8cm, DC = 5 cm Find: DB (Hint: Draw ABDF ⊥ )
R
Q P
S
T
300
16cm
A B
C D
E
M
N P
Q
B A
C
D
E
-
SUMMARY
Use the measurements as shown
32) to find x.
33) to find x and y. Exercise 31
Exercise 32 Exercise 33
2
2
x 450
450
2 x
y
600
300
-
5.4 Laws of Sine and Cosine
Outline: Sine, cosine, tangent, cosecant, secant, and cotangent
Laws of Sine
Law of Cosine
1) Sine, cosine, tangent, cosecant, secant, and cotangent
a) Definition:
Let PRQ∆ be a right triangle. Then we define six ratios as following
hypopp
RPRQθsin ==
opphyp
RQRP
sinθ1θ csc ===
hypadj
RPPQθ cos ==
adjhyp
PQRP
θ cos1θ sec ===
adjopp
PQRQθtan ==
oppadj
RQPQ
θtan 1θcot ===
b) Examples
Example 1: Find sin θ, sec θ, and tan θ
Solution
By Pythagorean Theorem, we have
RP = hyp = 29 x+
Thus,
29hypoppθsin
xx+
== ; 3
x9adjhypθ sec
2+== ; and
3adjoppθtan x== .
b) Inverse properties
i) For22παπ ≤≤− ,
=⇔= −
RPRQsin
RPRQsin 1αα
ii) sinβsinα then β,α If ==
R
Q P
hyp opp
θ
adj
R
Q P
3
θ
x
-
Example 2
Find the radius of the earth if α = 22.780 and h = 335 mi.
Solution
hrr
hypadjα cos
+==
Or
335rr22.78 cos 0
+=
335rr922.0
+=
≈−
=0.9221
5)(0.922)(33r 3960mi
-
Example 3: At the final approach, the shuttle flies at the altitude of 3,300ft and its the ground distance is 8,200ft from the landing strip. What is the glide angle for the shuttle to touch down at the beginning of the landing strip?
200,8300,3tanθ = ⇒ 01 22
8233tanθ ≈
= −
Example 4: Once upon the time, the Princess Rapunzel used her hair as the rope for a knight to climb up to the tower and rescue her from Mother Gothelf. How long was her hair?
50015tan 0 x= ⇒ x = (tan 150)(500) = 134 ft. Thus, the length of her hair was 143 + 8 = 151 ft.
Example 5:
Find the area of a scalene triangle.
Solution:
The area of triangle: bh21A = . Aslo,
ahsin =α ⇒=⇒ ha αsin αsin a b
21A ⋅=
3,300ft
8,200ft
θ
a
α
h
C B b
A
D
-
2) Law of sine
Let PRQ∆ be a scalene triangle and a, b, and c be the three sides that are opposite to the three angles γand β, α, respectively. Then
c
sinγb
sinβa
sinα==
Proof:
By example 4, the area of PRQ∆ is given by
αsin c b21A ⋅= βsin ca
21
⋅= = sin γ b a21
⋅= .
Divide by abc21
, we obtain c
sinγb
sinβa
sinα==
Example 7: In order to measure the distance across the Grand Canyon from A to C, a 1mi baseline AB is set up along the southern rim of the canyon. Use the surveyor to measure the interior angles of Δ ABC and apply the law of sine to find the distance. If ∠ BAC = 118.10 and∠ ABC = 58.10, find AC.
Solution:
m∠ BCA = 1800 – (118.10 + 58.10) = 3.80
⇒=AC
1.58sin1
8.3sin≈=
sin3.8sin58.1AC 13 miles
α β
ϒ
a b
c R
P
Q
-
Example 8:
A 12cm rod joins a piston to a 4.2 cm crankshaft (see the picture). If the rod makes an angle of 80, what is the distance d?
Solution:
By law if sine
⇒=4.2
8sin12
sinα⇒=
4.2)12)(8(sinsinα ≈
= −
4.2)12)(8(sinsin 1α 23.40
ϒ = 180 – ( 8 + 23.4) = 148.60 .
⇒=4.2
8sin6.148sind
≈=8sin
(4.2)(sin148.6)d 15.7 cm
-
3) Law of cosine
Let PRQ∆ be a scalene triangle and a, b, and c be the three sides that are opposite to the three angles γand β, α, respectively. Then
cosα cb2cba 222 ⋅⋅−+=
cosβ ca2cab 222 ⋅⋅−+=
cosγ ba2bac 222 ⋅⋅−+=
Proof
Given: PQR∆ with PQ = a; PR = b; RQ = c
Prove: cosα cb2cba 222 ⋅⋅−+=
STATEMENTS REASONS
1) PQR∆ with PQ = a; PR = b; RQ = c 1) Given
2) Draw RQPH ⊥ ; RH = x; PH = y ; HQ= c – x
2) Construction
3) 222 )( xcya −+= (i)
222 yxb += (ii)
3) Pythagorean Theorem
4) 2222 )( xcxba −+−= 4) Substitution (ii) into (i)
5) cxcba 2222 −+= 5) Simplify
6) bxcos =α
6) Definition of cosine
7) x = b cosα 7) Cross Multiply
8) cosα cb2cba 222 ⋅⋅−+= 8) Substitution
α β
ϒ
a b
c – x R
P
Q x
y
H
-
Example 9:
Two satellites are put in the geostationary orbit for the communication between the space shuttle and the White Sand Missile Range, New Mexico. Each satellite is 22,300 mi above the earth, and they are 1300 apart. If the radius of the earth is about 3964 mi, how far is the satellite from the White Sand Missile Range?
Solution:
SC = 22300 +3964 = 26264
WS = 022 65cos26264*3964*2262643964 −+ = 23,910 mi
-
Example 10: The geophysicist use sound to search for oil or examining fault lines. The dynamite is detonated at a point P and transmitted sound signals to the point Q, a D distance from P. It takes T1 seconds for the first signal to arrive Q after traveling along the surface with the speed c1; it takes T2 seconds for the second signal to arrive Q passing the midpoint O with the speed c2; it takes T3 seconds for the third signal to arrive Q passing R, O, and S with the same speed c2. If c1 < 2c , show that T2 is least time among them.
Solution:
c 2 > c1 31 TT >⇒
m PRO ∠ = θ + 900 cosθ θ)cos(90PRO)cos( −=+=∠⇒
By law of cosine, we have
PRO)os(2(PR)(RO)cROPRPO 222 ∠−+=
osθ2(PR)(RO)cROPRPO 222 ++=
Also, PRhcosθ =
Thus,
2h(RO)ROPRPO 222 ++=
Since h < PR, then
( )222222 ROPR2(PR)(RO)ROPR2h(RO)ROPRPO +=++>⇒>
-
Example10 (Napoleon Triangles)
On each side of a scalene triangle ABC, draw an equilateral triangle outside (or inside) of ∆ ABC. The triangle MNP that is formed by connecting three centroids of three equilateral triangles is also an equilateral triangle.
Given: ∆ABC with AC = a; AB = b; BC = c; M, N, P are centroids
∆ ABE, ∆BCF, ∆CAD are equilateral ∆’s Prove: ∆MNP is equilateral∆
A
B C
M
N
P
E
D
F
a b
c
t
s
-
Given: ∆ABC with AC = a; AB = b; BC = c; M, N, P are centroids
∆ ABE, ∆BCF, ∆CAD are equilateral ∆’s Prove: ∆MNP is equilateral∆
STATEMENTS REASONS
1) ∆ABC with AC = a; AB = b; BC = c; M, N, P centroids 1) Given
2) s = �23� �𝑎𝑎√3
2� = 𝑎𝑎
√3; t = �2
3� �𝑏𝑏√3
2� = 𝑏𝑏
√3; 2) Centroid is 2/3 the length of the median
3) 𝑀𝑀𝑀𝑀2 = 𝑠𝑠2 + 𝑡𝑡2 − 2𝑠𝑠𝑡𝑡𝑠𝑠𝑠𝑠𝑠𝑠(𝐴𝐴 + 600)
3) Law of cosine
4) 3𝑀𝑀𝑀𝑀2 = 𝑎𝑎2 + 𝑏𝑏2 − 2𝑎𝑎𝑏𝑏𝑠𝑠𝑠𝑠𝑠𝑠(𝐴𝐴 + 600) 4) Substitution (2) into (3) and simplification
5) 3𝑀𝑀𝑁𝑁2 = 𝑏𝑏2 + 𝑠𝑠2 − 2𝑏𝑏𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠(𝐵𝐵 + 600) 5) Repeat from(2) to (4) for NP
6) ∆ ABE, ∆BCF, ∆CAD equilateral ∆’s 6) Given
7) a = b = c; m∠A = m∠B = m∠C 7) Sides and angles of equilateral ∆ are ≅
8) 3𝑀𝑀𝑁𝑁2 = 𝑎𝑎2 + 𝑏𝑏2 − 2𝑎𝑎𝑏𝑏𝑠𝑠𝑠𝑠𝑠𝑠(𝐴𝐴 + 600) 8) Substitution (7) into (5)
9) 3𝑀𝑀𝑀𝑀2 = 3𝑀𝑀𝑁𝑁2 9) Transition
10)MN = NP 10) Divide by 3 and square root
11) NP = PM 11) Repeat from (2) to (10) for PM
12) MN= NP = PM 12) Transition
13) ∆MNP equalateral 13) Definition of equilateral ∆
-
Homework 5.4
1) Find sine, cosine, and tangent of the angle Ɵ
2) Find sine, cosine, and tangent of the angle α
3) Find sine, cosine, and tangent of the angle β
Exercise 1
Exercise 2 Exercise 3
Use the law of sine or law of cosine to solve each triangle
4)β = 1320, ɣ = 170, b = 67ft
5) α = 570, ɣ = 1120 , c = 24ft
6) β = 660, a = 14 m, c = 20m
7) ɣ = 540, a = 112ft, b = 87ft
8) β = 720, α = 370, ɣ = 710
9) a = 15ft, b = 18ft, c = 22ft
10) β = 123, ɣ = 22, a = 108cm
11) An 8 yard ladder is against the house as indicated in the figure. How far will the ladder reach up the house?
12) A boat is approaching a 200ft high vertical cliff as in the figure. Find the distance d if the angle Ɵ = 250.
R
Q P
5
θ
x
α
R
Q P
x
θ
3
α
x
α
R
Q P
3
β
4
x
α β
ϒ
a b
c R
P
Q
-
13) The military often uses the air-to-air refuel for some large jets for lacking of landing facilities. If the angle of elevation Ɵ = 350, the vertical distance between two jets a = 184.4ft, the length c of the hose.
14) The military often uses the air-to-air refuel for some large jets for lacking of landing facilities. If the angle of elevation Ɵ = 320, the horizontal distance between two jets b = 120ft, the length c of the hose.
15) Use the information in the figure to find the distance d.
16) Use the information in the figure to find the distances x and y.
Exercise 8 Exercise 9
17) To accommodate cars of most sizes, a parallelogram parking space needs to contain a 18ft x 8ft rectangle as indicated in the figure. If the angle of the parallelogram Ɵ = 720, find all sides of the parallelogram.
18) To accommodate cars of most sizes, a parallelogram parking space needs to contain a 18ft x 8ft rectangle as indicated in the figure. If the angle of the parallelogram Ɵ = 680, find all sides of the parallelogram.
19) In order to measure the distance across the Grand Canyon from A to C, a 1mi baseline AB is set up along the southern rim of the canyon. Use the surveyor to measure the interior angles of Δ ABC and apply the law of sine to find the distance. If ∠ BAC = 28.50 and∠ ABC = 144.60, find AC.
-
20) A 12cm rod joins a piston to a 4.2 cm crankshaft (see the picture). If the rod makes an angle of 80, what is the distance d?
21) To get the distress swimmer, the lifeguard must run some distance along the beach at the rate of 5m/s, enter the water at the rate of 1.7m/s. If d = 380m, c = 76m, and Ɵ = 510, find the distance of running and the distance of swimming.
22) An island is 4mi offshore in a large bay. A water pipeline is to be run from a water tank on the shore to the island with angle Ɵ = 150, as shown in the figure. The pipeline costs $40,000 per mile in the ocean and $20,000 per mile on the land. Use the measurements as shown in the figure to find the total cost of the pipelines.
23) The underwater cable runs cross the lake from point A to point B. If the distance AC = 112 yards and m∠ ACB = 1180 and m∠ ABC = 190. Find distance AB.
24) A tree growing on the hillside cast a157ft shadow straight down the hill. If the hill slope is 110 and the angle of elevation of the sun is 420, find the height of the tree.
25) Find the height of the tree in problem 17) if the shadow length is 102ft, the hill slope is 150, and the angle of elevation is 620.
-
26) An airplane flying on the level course directly toward a beacon on the ground with two angles of depression as indicated in the figure. Find the altitude h of the plane.
27) Repeat the problem 19) with angles of depression of 310 and 550.
28) Two guy wires are attached to a radio tower as shown in the figure. If x = 50ft, find the angle Ɵ.
29) Two guy wires are attached to a radio tower as shown in the figure. If the angle Ɵ = 200, find x.
30) To measure the length of the lake BC, a surveyor measures AC = 89 m and AC = 74 m, and ∠ BAC = 950. Find the length BC.
-
31) A tunnel is to be constructed through a mountain to connect two reservoirs. Use the information in the figure to find the length of the tunnel.
32) Repeat the problem 24) with angles of depression of 400 and 380.
33) A fire lookout post is to be constructed as indicated in the figure. If support poles AD = BC = 18ft, the distance between the top two support poles DC = 12ft, and the poles tilt 80 from the vertical line, find the brace AC and the distance AB between two poles on the ground level.
34) Repeat the problem 26) with support poles tilt 110 from the vertical line.
35) A satellite S is sighted by a tracking station T. If TS = 1,034mi, the angle of elevation above the horizontal line is 320, and the radius of earth = 3,964mi, how high is the satellite above the earth?
36) Repeat the problem 28) with the angle of elevation above the horizontal line is 250.
37) Los Angeles and New York are 2,451miles a part. After the pilot flies 560 miles from Los Angles, she realizes that the plane is 150 off the course to New York. How far is the plane from New York at this time?
-
38) Los Angeles and San Francisco are 600kms a part. After the pilot flies 200km from Los Angles, she realizes that the plane is 200 off the course to San Francisco. How far is the plane from San Francisco at this time?
Summary
39) What are definitions of sine, cosine, and tangent?
31) State the law of sine.
32) State the law of cosine.
-
Chapter 5 Exam
1) Given: PQRS is a parallelogram.
Prove: TSR∆ ~ RQU∆
2) The segments QS and QT trisect the rt.∠ A. Find RT, SP, and PR.
3) A tree growing on the hillside cast a157ft shadow straight down the hill. If the hill slope is 110 and the angle of elevation of the sun is 420, find the height of the tree.
4) The building is 30ft tall and casts a shadow 8ft wide in front of the building. If Mindy is 5’ 6” tall, how far away from the building can she stand and still be able in the shade?
P
S R
Q
T
U
R
Q P
S
T 600
6cm
30’
8’
-
5) Given: ABC is right triangle with angle A is right angle. AB = b, BC = c and AC = a
Prove: a2 + b2 = c2
6) Given: DF||AB and AF||CE
Prove: AGDECEAB ⋅=⋅
7)____BD bisects ABC∠ , BC = 2x, DC = x, AD = x + 1, and AB = 3x – 1. Find
BC.
A
B C
a b
c
B
A C D
E
F
G
A
2x B
C
D
x
x+1
3x−1
-
Chapter 6 Circles 6.1 Introduction to circles
6.2 More Chords of circles
6.3 More Angles of circles.
-
6.1 Introduction to circles
Outline: Definition
Chords, Angles, and Arcs.
1) Definitions
a) A circle is a collection of points whose distances from a fixed point equal to a constant r. The fixed point is the center; the constant r is the radius.
b) Congruent circles are circles with congruent radii.
c) Concentric circles are circles with a common center.
2) Chords, Angles, and Arcs
a) Chords
A chord PR is a line segment connecting two points on the circle.
Diameter PQ is a chord containing the center of the circle.
Tangent PT is a line that intersects the circle at one point.
Secant TQ is a line that intersects the circle at two points.
.
r
. O
P
Q
. O
P R
T
-
Theorem 1
In a circle, congruent chords have congruent arcs.
Given: CDAB ≅
Prove: CD AB ≅
Proof:
STATEMENTS REASONS
1) CDAB ≅ 1) Given
2) Draw OA , OB , OC , and OD 2) Construction
3) ≅≅ OBOA ODOC ≅ 3) All radii are ≅
4) OCDΔOAB ∆≅ 4) SSS
5) 21 ∠≅∠ 5)CPCTC
6) 2m1m ∠=∠ 6) Definition of ≅ s'∠
7) CD mAB m =
7) Measure of central∠ of a circle = measure of intercepted arc
8) CD AB ≅
8) Definition of ≅ arc
Theorem 2
In a circle, congruent arcs have congruent chords.
The proof is left as the exercise
2
D
C
A
B
O 1
-
Theorem 3
Chords with the same distance from the center of the circle are congruent.
Given: ABOE ⊥ ; CDOF ⊥ ; OFOE ≅
Prove: CDAB ≅
STATEMENTS REASONS
1) ABOE ⊥ ; CDOF ⊥ ; OFOE ≅ 1) Given
2) Draw OA , OB , OC , and OD 2) Construction
3) ≅≅ OBOA ODOC ≅ 3) All radii are ≅
4) OCFΔOAE ∆≅ 4) HL
5) CFAE ≅ 5)CPCTC
6) AE = CF 6) Definition of ≅ segments
7) 2(AE) = 2(CF) 7) Multiply by 2
8) OCD and ΔOAB ∆ isosceles 8) Definition of isosceles Δ
9) BEAE = ; DFCF = 9) In an isosceles Δ , an altitude is also ⊥ bisector.
10) EBAE AB += ; FDCF CD += 10) Segment addition
11) AB= 2(AE); CD = 2(CF) 11) Substitution 8) into 9) and simplify
12) AB = CD 12) Transition
13) CDAB ≅ 13) Definition of ≅ segments
F
B
A
D
O
E
C
-
Theorem 4
Congruent chords have the same distance from the center of the circle.
The proof is left as the exercise
b) Angles
i) Definition: A central angle POQ∠ is an angle whose sides are radii and its vertex is at the center of the circle.
ii) Postulate
The measure of a central angle of a circle is equal to the measure of its intercepted arc.
iii) An inscribed angle PTQ∠ or QTR∠ is an angle whose sides are two chords or a chord and a tangent and its vertex is a point on the circle or a tangent point.
iv) Theorem 1
The measure of an inscribed angle of a circle is one half the measure of its intercepted arc.
Proof:
Case 1: Inscribed angle formed by a chord and a diameter.
Given: AB is diameter; CD is a chord Prove: AOCmABCm ∠=∠
Q
. O
P
T
R
O
A
B
C
1
2
-
STATEMENTS REASONS
1) AB diameter; CD chord 1) Given
2) Draw OC 2) Construction
3) OCOA ≅ 3) All radii are ≅
4) ΔOBC isosceles 4) Definition of isosceles Δ
5) 21 ∠≅∠ 5)Base s'∠ of isosceles Δ are≅
6) 2m1m ∠=∠ 6) Definition of ≅ s'∠
7) AOCm2m1m ∠=∠+∠ 7) Measure of ext. ∠ of a Δ = sum of measures of two non- adjacent s'∠
8) AOCm2)2(m ∠=∠ 8) Substitution 5) into 6) and simplify
9) AOCm212m ∠=∠
9) Divide
10) ACmAOCm ∠=∠ 10) Postulate section 6.1
11) ACm212m ∠=∠
11) Substitution 9) into 8)
Case 2: The center of circle is inside of inscribed angle. Case 3: The center of the circle is outside of inscribed angle.
Proofs are left as exercises
.O A
B
C
A
.O
B
C
-
v) Corollary: The angle inscribed in a semi-circle is a right angle.
The proof is left as an exercise.
Example: Huygens’ Principle.
In figure, the production of Huygens’ wavelets, secondary waves, propagates outward with the same speed c of the first wave. At the new waterfront, two wavelets propagate from P to T and from Q to S with the same speed c. Since the distance equals to product of speed c and Δt . Thus, PT = SQ. Now we can show that the incident angle θ1 is congruent to reflective angle θ2.
Given: S and T on the semi-circle center O; PT = SQ
Prove: 21 θθ ≅
STATEMENTS REASONS
1) S and T on the semi-circle center O 1) Given
2) ∠ S and ∠ T are rt s'∠ 2) Angles inscribed in a semi-circle is rt s'∠
3) ΔPQS and ΔPQT are rt. sΔ' 3) Definition of rt. sΔ'
4) PQPQ ≅ 4) Identity
5) PT = SQ 5) Given
6) SQPT ≅ 6) Definition of ≅ segment
7) ΔPQSΔPQT ≅ 7) HL
8) 21 θθ ≅ 8)CSSTP
P
Q
S
T
Θ2 Θ1
S T
P Q
P
New wave-front
O Q A’ A
T S
Θ2 Θ1
-
vi) Theorem 2
If two inscribed angles intercept the same arc, these angles are congruent.
The proof is left as an exercise.
c) Arcs
i) Definition: An arc, denoted AB , is a part of a circumference that is from point A to point B.
ii) Arc Addition Postulate: m AB + m BC = m ABC
d) The relation between angle and arc
It follows from the measure of the central angle is equals to the measure of its intercepted arc,
we have the following proportion: 𝜃𝜃3600
= 𝑙𝑙𝐶𝐶
Example : Radius of the earth by Eratosthenes (276 – 194 BC)
When the sun was directly over the city Syene, the angleα at the Alexandria would be α = 7.20. The distance l between Syene and Alexandria was about 5,000 stadia (1 stadium )ft 516.73≈ . Let C be the circumference of the earth. Then
⇒=C
50003607.2
0
0
⇒≈= dia.250,000sta2.7
)360)(5000(C ia397.89stadR ≈ .
It’s about 3925 mi, which is 35 mi less than actual radius.
B
O
A
C
l
Ɵ
O
-
Homework 6.1 Given: m∠ MOQ = 750 and m ∠QOP = 850, QN is a diameter. Find 1) m MN and m NP 2) m ∠MON and m ∠ NMP 3) m ∠NQP and m ∠MNQ Given: 𝐴𝐴𝐵𝐵||𝐶𝐶𝐶𝐶, 𝐴𝐴𝐵𝐵 ≅ 𝐶𝐶𝐶𝐶 , and m ∠ ACD = 32O . Find 4) m ∠BDC, m∠ BAC, m ∠ABD 5) m∠ AOB 6) m DC 7)m AD Exercise 1 – 4 Exercise 4 – 7 8) If PQ = 30” and QR = 15”. Find PR. 9) If AB = 4cm, BC = 14 cm, DC = 8cm. Find AD. 10) If m AB = 40o, m BC = 120o, and m CD = 80o, find m ∠ ABC Exercise 8 Exercise 9 and 10
S
M
P
N
O Q
75o
85o
C
D
B
A
O
P
Q
R O A
B
C
D
O
-
11) Find m ∠ A of the American Captain Shield logo. 12) If m ∠ N = 50o and m PQ = 2
3 MN, find m ∠P
Exercise 11 Exercise 12 13) If a wind turbine has 3 equally spaced blades, find the angle formed by two consecutive blades. 14) If a fan has 5 equally spaced blades, find the angle formed by two consecutive blades.
If m PQ: m QM: m MP = 2: 3 :4. Find 16) m ∠ 1 17) m ∠ 2 18) m ∠ 3 Exercise 16 – 18 Given: 𝑀𝑀𝑀𝑀 is a diameter, PN = ON.
Find
19) m PN
20) m MP
21) m ∠ N Exercise 19 – 21
1 O
M Q
P
2
3
O
M
P
Q
N
A
B
C D
E
P
N
M O
-
22) Prove that in a circle, congruent arcs have congruent chords.
Given: CD AB ≅
Prove: CDAB ≅
STATEMENTS REASONS
1) Draw AB , CD , OA , OB , OC , and OD 1) ?
2) CD AB ≅
2) ?
3) CD mAB m =
3) ?
4) 2m1m ∠=∠ 4) ?
5) 21 ∠≅∠ 5) ?
6) ≅≅ OBOA ODOC ≅ 6) ?
7) OCDΔOAB ∆≅ 7) ?
8) CDAB ≅ 8) ?
2
D
C
A
B
O 1
-
23) Given: 𝐴𝐴𝐵𝐵||𝐶𝐶𝐶𝐶 . Prove: ABCD is an isosceles trapezoid.
24) Congruent chords have the same distance from the center of the circle.
Given: ABOE ⊥ ; CDOF ⊥ ; CDAB ≅ Prove: OFOE ≅
STATEMENTS REASONS
1) 1) Given
2) Draw OA , OB , OC , and OD 2)
3) ≅≅ OBOA ODOC ≅ 3)
4) 8) ∆𝑂𝑂𝐴𝐴𝐵𝐵 ≅ ∆𝑂𝑂𝐶𝐶𝐶𝐶 4) SSS
5) ∠ 3 ≅ ∠ 4 5)CPCTC
6) 6)⊥ lines form rt ∠ ’s
7) 21 ∠≅∠ 7)
8) ∆𝑂𝑂𝐴𝐴𝑂𝑂 ≅ ∆𝑂𝑂𝐶𝐶𝑂𝑂 8)
9) 9)
D C
O
B A
F
B
A
D
O
E
C
-
25) Give: Diameters 𝑁𝑁𝑃𝑃 and 𝑀𝑀𝑀𝑀 ; 𝑀𝑀𝑃𝑃; 𝑁𝑁𝑀𝑀; and 𝑀𝑀𝑁𝑁 Chords Prove: ∆𝑀𝑀𝑀𝑀𝑁𝑁 ≅ ∆𝑃𝑃𝑁𝑁𝑀𝑀
26) Given: Chords 𝑀𝑀𝑁𝑁 , 𝑀𝑀𝑀𝑀, 𝑀𝑀𝑃𝑃 and 𝑃𝑃𝑁𝑁 Prove: ∆𝑀𝑀𝑀𝑀𝑀𝑀~∆𝑃𝑃𝑀𝑀
27) Five points are equaly spaced on a circle. A pentagram, a five- pointed star, is formed by joining two non-consecutive points at a time. Find the measure of the interior acite angle and the measure of the interior reflex angle.
28) Given: 𝐴𝐴𝐵𝐵 is a diameter, AC = 6” and AB = 10”; 𝐴𝐴𝐶𝐶 ≅ 𝐴𝐴𝐶𝐶 ; Find CD
O M
P
Q
N
R
Q M
N P
Interior acute
Interior reflex
B
A C
D
O
-
29) Reno, Nevada is approximately due north of Los Angeles. If the latitude of Reno is 40° N and latitude of Los Angeles is 34° and the radius of Earth is about 3960 miles, find the north-south distance between the two cities. 30) If San Francisco, CA is 37° 50’N and Seattle, WA is 47°40’N and the radius of Earth is about 3960 miles, find distance between the two cities. 31) If Dallas, TX is 32° 50’N and Lincoln, NE is 40°50’N and the radius of Earth is about 3960 miles, find distance between the two cities. 32) If Phoenix, AZ is 33° 30’N and Salt Lake city, UT is 40°40’N and the radius of Earth is about 3960 miles, find distance between the two cities. Prove that the measure of an inscribed angle of a circle is one half the measure of its intercepted arc. 33) Case 2: The center of circle is inside of inscribed angle.
34) Case 3: The center of the circle is outside of inscribed angle.
Case2 Case 3 35) Prove the corollary: The angle inscribed in a semi-circle is a right angle. 36) Prove that if two inscribed angles intercept the same arc, these angles are congruent.
SUMMARY:
37) The measure of an _____________angle of a circle is one half the measure of its _________ ______.
38) The measure of a ________ angle of a circle is equal to the measure of its ____________ ____.
39) If two _______ angles intercept the _______ arc, these angles are _______.
40) The angle inscribed in a ______-circle is a ______ angle.
41) In a circle, congruent ______ have congruent ______
42) Congruent _______ have the same __________ from the center of the circle.
43) Chords with the same ________ from the center of the circle are ____________.
44) State the proportion between angle and arc.
.O A
B
C A
.O
B
C
-
6.2 More Chords of Circles
Outline: Theorems
Theorems
1) Theorem 1: If a chord is perpendicular to the radius, the radius bisects the chord.
Given: ABOC ⊥
Prove: MBMA ≅
STATEMENTS REASONS
1) ABOC ⊥ 1) Given
2) Draw OA and OB 2) Construction
3) OBOA ≅ 3) All radii are ≅
4) ΔOBC isosceles 4) Definition of isosceles Δ
5) MBMA ≅ 5) In an isosceles triangle, an altitude is also the median.
Example 2: If the Hyperloop pod has the radius of 5ft, and the height of 9ft, what is the width of track?
Since the radius perpendicular to the chord, it bisects the chord. Thus, by Pythagorean Theorem,
we have AB = 2 MB = 2 1625 − = 6ft
M
O
B A
C
O
B A
5ft 4ft
-
2) Theorem 2
If a radius bisects the chord, the radius is perpendicular to the chord
The proof is left as the exercise.
3) Example 3: A stack of bubble.
A 3-in radius hemispherical bubble is place on a 5 in radius spherical bubble. Another 1 in radius hemisphere is then placed on the second hemisphere (see figure). Find the height FO of the bubble tower.
Solution:
Since AB is the diameter of the hemisphere, then QA = QB = QC = 3 in
Since the radius OE bisects the chord AB , then ABOE ⊥ .
By Pythagorean Theorem, we have 22 ACOAOC −= = 4 in.
Similarly, 22GECGCE 22 =−= in
OF = OC + CE + EF = 4 + 22 + 1 = 5 + 22 in
Example 4:
The ramp that is used in the Olympic skateboard events or BMX competitions is in the shape of an inversed cycloid, called a brachistochrone or a “half pipe”. The half pipe curve allows a snowboarder to get from A to D in the shortest time of any curve including the line segment AD. If the radius of the curvature is defined as the radius of a circle containing the arc ADC, use the dimension shown to find the radius of curvature r
Solution:
Since triangle OAB is a right triangle, then by Pythagorean Theorem, we can approximate the radius of the curvature. That is 222 28.9812)(rr +−= . Hence, 41ftr ≈
G
O
D
E
F
5 in
3 in A B C
1in
12ft 58ft 28.98ft
r
D
28.98ft 28.98ft
12ft
O
A B C
-
4) Theorem 3
The tangent is perpendicular to the radius at the tangent point.
Given: Circle center at O; the tangent TQ of circle O at tangent point Q.
Prove: ↔
⊥ TQOQ at Q Proof by contradiction
STATEMENTS REASONS
1) Circle O; tangent T at the tangent point Q. 1) Given
2) ↔
⊥ TQOQ 2) Assumption
3) Draw↔
⊥ TQOS 3) Construction
4) S lies on the line TQ, not on the circle 4) Q is a tangent point and OQOS ≠
5) OS > OQ 5) S lies on the exterior of the circle.
6) OSQ∆ is a rt. ∆ with rt. S∠ and OQ hypotenuse 6) Definition of a rt. ∆
7) OS < OQ, a contradiction to 5) 7) Hypotenuse is the longest side of a rt. ∆
8) ↔
⊥ TQOQ at Q 8) By contradiction proof
5)Theorem 4: The measure of a special inscribed angle, an angle formed by a chord and a tangent intersecting at the tangent point, is one-half the measure of its intercepted arc.
Proof is left as an exercise.
O
Q S T
O
Q T
-
6) Theorem 5
Two tangents to a circle from an exterior point are congruent.
Proof
Given: QA and QB tangents to circle O
Prove: QBQA ≅
STATEMENTS REASONS
1) QA and QB tangents 1) Given
2) Draw AB 2) Construction
3)
ABm21QBAm ∠=∠ ;
ABm21QABm ∠=∠
3) The measure of a special inscribed angle of a circle is one half the measure of its intercepted arc.
4) ABmQBAm Q∠=∠ 4) Transition
5) ABQBA Q∠≅∠ 5) Definition of ≅ s'∠
6) QBQA ≅ 6) In a ∆ , the opp. sides to ≅ s'∠ are≅
Q
B
A
-
Example 4:
Packing two identical 2cm-circles and a smaller circle radius r into a larger 4cm-circle so that none of circles overlap. Find the radius r of the smaller is as large as possible.
Solution:
By the definition of tangent, we have
21
42tan ==θ So Θ = 01 57.26
21tan ≈
− ;
By the Pythagorean Theorem, we get
5224OP 22 =+= ;
Apply the law of cosine, we obtain
( ) ( ) ( ) )57.26cos(522522 222 rrr −+=+ .
Thus, r = 33.157.26cos51
4≈
+cm
r +2
θ
4cm
2cm O Q
P
S
O
P
Q 2 cm
r
S
-
5) Example 5 (Bow waves) If the boat is at the point A, it produces a bow wave that radiates out in a circle with radius r. By the time the boat reaches the point B, the radius of the bow wave’s circle will radiate a distance of R (see the figure)
Since the tangent is perpendicular to the radius at the tangent point, ΔABC is a right triangle.
By theorem 4, BDBC ≅ . This implies AB is the angle bisector of CBD ( or angle θ.) Why?
Also, the distance R = Vwt, where Vw is the speed of the wave and t is the time.
Similarly, the distance AB = Vbt, where Vb is the speed of the boat and t is the time.
By the definition of sine, we obtain
𝑠𝑠𝑠𝑠𝑠𝑠 �𝜃𝜃2
� = 𝑅𝑅𝐴𝐴𝐴𝐴
= 𝑉𝑉𝑤𝑤 𝑡𝑡𝑉𝑉𝑏𝑏 𝑡𝑡
, or 𝑠𝑠𝑠𝑠𝑠𝑠 �𝜃𝜃2
� = 𝑉𝑉𝑤𝑤𝑉𝑉𝑏𝑏
If a boat travels at a constant speed, it generates a bow wave whose speed is slower than the speed of the boat. If a speed boat travels at 45mph and the angle of the bow wave is 720, find the speed of the bow wave.
Solution
𝑠𝑠𝑠𝑠𝑠𝑠 �720
2� = 𝑉𝑉𝑤𝑤
45⇒ Vw = (45)(sin 360 )≅ 26.45 mph
C
A B
R
r
D
-
6) Sonic Cone:
Similarly, if an aircraft is flying faster than the speed of sound, it generates a sound wave whose speed is slower than the speed of the airplane. Thus, we have
a
s
SS
2θsin = , where Ss is the speed of the sound wave, Sa is the speed of the airplane.
7) Nuclear particles:
In 1958, Cerenkov, Frank, and Tamm received a Nobel Prize for their work based on the bow wave analysis. Since nuclear particles can be made to move faster than the speed of light (6.71x 108 mph), the three physicists was able to determine the speed of nuclear particle by measuring the light cone angle θ and using the formula
p
l
SS
2θsin = , where Sl is the speed of the light wave, Sp is the speed of the particle.
If a high-energy nuclear particle passes thru the crown glass with the light cone angle of 600, find the speed of the particle.
≈⋅
= 08
p sin30106.71S 12.34 x 108mph.
light cone Glass
Nuclear particle
-
HOMEWORK 6.2 Given: Concentric circles with center O. AB = 8 cm; EB = 2 cm; AB ⊥ 𝑂𝑂𝐶𝐶 Find: 1) OE 2) OC Exercise 1 and 2 Given the circle center O with radius 15 cm, m∠AOB = 900, and 𝐴𝐴𝐵𝐵 ⊥ 𝑂𝑂𝐶𝐶. Find 3) OD 4) AB 5) DE Exercise 3 – 5 Exercise 6 – 8 Given the circle center O with radius 20 cm, m∠AOB = 1200, and 𝐴𝐴𝐵𝐵 ⊥ 𝑂𝑂𝐶𝐶. Find 6) OD 7) AB 8) DE Given: 𝑀𝑀𝑀𝑀 is a diameter, 𝑁𝑁𝑀𝑀 is a tangent, 𝑀𝑀𝑀𝑀 is a chord, and 𝑚𝑚 𝑀𝑀𝑀𝑀� = 250 Find 9) m∠ PNM 10) m∠MNR 11) m∠ PNR Given: E is the midpoint of 𝐴𝐴𝐵𝐵� , m∠DAE = 150 and OD = 6. Find 12) OA 13) DE Exercise 9 – 11 Exercise 12 and 13
A B
C
D
O
E
A B
O
D
E
A B
O
D
E
O
M
N
P
R
A B
O
D
E
-
14) Given: 𝑂𝑂𝑃𝑃 is the common chord of circle O and circle Q. Prove: 𝐴𝐴𝐶𝐶 = 1
2𝐴𝐴𝐵𝐵
STATEMENTS REASONS
1) Draw OA , OB ,𝑃𝑃𝐴𝐴 , and 𝑃𝑃𝐵𝐵 1) ?
2) 𝑂𝑂𝐴𝐴 ≅ 𝑂𝑂𝐵𝐵 ; 𝑃𝑃𝐴𝐴 ≅ 𝑃𝑃𝐵𝐵
2) ?
3) OAQB is a kite
3) ?
4) 𝑂𝑂𝑃𝑃 ⊥ 𝐴𝐴𝐵𝐵 4) ?
5) 𝐴𝐴𝐶𝐶 = 12
𝐴𝐴𝐵𝐵 5) ?
15) Given: 𝑂𝑂𝑃𝑃 is the common chord of circle O with radius 5 cm and circle Q with radius 5 cm. AB = 12 cm. Find OQ
Q
A
B
O
O Q
A
B
C
-
16) Given: 𝑂𝑂𝑃𝑃 is the common chord of circle O with radius 3 cm and circle Q with radius 5 cm. AB = 4 cm. Find OQ 17) Given: 𝑂𝑂𝑃𝑃 is the common chord of circle O and circle Q. 𝑀𝑀𝑀𝑀is a common tangent Prove: 𝑚𝑚∠𝑂𝑂𝐴𝐴𝑃𝑃 = 900 18) Given 𝐴𝐴𝐵𝐵� = 𝐶𝐶𝐶𝐶� Prove: 𝐴𝐴𝐶𝐶||𝐵𝐵𝐶𝐶 19) Given: 𝐴𝐴𝐶𝐶 𝑎𝑎𝑠𝑠𝑎𝑎 𝐵𝐵𝐶𝐶 are tangents to the circle center at O Prove: ∆ABC is isosceles Exercise 16 Exercise 17
O Q
A
B
O Q
A
B
M
N
A
C
B
D
C
B
A
-
20) Prove that if a radius bisects the chord, the radius is perpendicular to the chord Given: MBMA ≅
Prove: ABOC ⊥
STATEMENTS REASONS
1) 1) Given
2) 2)
3) OBOA ≅ 3) All radii are ≅
4) ΔOBC isosceles 4)
5) 5) In an isosceles triangle, the media is also the altitude.
21) Packing two identical 3cm-circles and a smaller circle radius r into a larger 6cm circle so that none of circles overlap. Find the radius r of the smaller is as large as possible.
O
B A
C
M
O
P
Q 3 cm
r
S
r +3
3√5
θ
6cm
3cm O Q
P
S
-
22) Find the largest radius of two identical circles that can be backed in to a unit circle. 23) Find the largest radius of three identical circles that can be backed in to a unit circle. Exercise 20 Exercise 21 24) Mindy is watching the seascape through her binocular from the veranda of a beachfront hotel, and suddenly a ship appears on the horizon. If Mindy is 80ft above the earth, how far is the ship out at sea? Given that 1 mi = 5280ft. 25) A passenger on an air plane is watching the seascape through his binocular, and suddenly a tsunami appears at the horizon. If the airplane is 4 mi above the earth, how far is the tsunami out at a sea? 26) If a speed boat is traveling at 121 km/hr and the angle between the bow waves is 740, how fast is the bow wave moving? 27) If a boat is traveling at 55 km/hr and the angle between the bow waves is 540, how fast is the bow wave moving? 28) Mach number is the ratio of the speed of an airplane Va to the speed of sound Vs in the surrounding medium. Mach 1 is the speed of an airplane flying at the speed of sound at sea level, approximate 750mph.The British-French commercial airplane Concorde flies at Mach2. Find the vertex angle Ɵ of the sound cone. 29) If the fighter jet F-35 flies at the maximum speed of Mach 1.6, find the vertex angle Ɵ of the sound cone. 30) In a crown glass, light travels at approximate 2 x 1010 cm/sec. If a high-energy particle passing through the glass creates the light cone of 700, how fast the light travel? 31) In a crown glass, light travels at approximate 2 x 1010 cm/sec. If a high-energy particle passing through the glass creates the light cone of 800, how fast the light travel?
4
3960
-
32) A 4-in radius hemispherical bubble is place on a 6-in radius spherical bubble. Another 2-in radius hemisphere is then placed on the second hemisphere (see figure). Find the height FO of the bubble tower.
33) Given: 𝐴𝐴𝐵𝐵 𝑎𝑎𝑠𝑠𝑎𝑎 𝐶𝐶𝐶𝐶are chords of circle cnter O, 𝑂𝑂𝑂𝑂𝑎𝑎𝑠𝑠𝑎𝑎 𝐺𝐺𝐺𝐺are the perpendicular bisectors of 𝐴𝐴𝐵𝐵 𝑎𝑎𝑠𝑠𝑎𝑎 𝐶𝐶𝐶𝐶 Prove: The center O is the intersection point of 𝑂𝑂𝑂𝑂𝑎𝑎𝑠𝑠𝑎𝑎 𝐺𝐺𝐺𝐺
STATEMENTS REASONS
1) 𝐴𝐴𝐵𝐵 𝑎𝑎𝑠𝑠𝑎𝑎 𝐶𝐶𝐶𝐶are chords of circle O 𝑂𝑂𝑂𝑂𝑎𝑎𝑠𝑠𝑎𝑎 𝐺𝐺𝐺𝐺 are ⊥bisectors of 𝐴𝐴𝐵𝐵 𝑎𝑎𝑠𝑠𝑎𝑎 𝐶𝐶𝐶𝐶
1)
2) Draw 𝑂𝑂𝐴𝐴, 𝑂𝑂𝐵𝐵,𝑂𝑂𝐶𝐶 , and 𝑂𝑂𝐶𝐶 2)
3) OBOA ≅ ; 𝑂𝑂𝐶𝐶 ≅ 𝑂𝑂𝐶𝐶 3)
4) O is equal distance from A and B O is equal distance from A and B
4)
5) O is a point of perpendicular bisectors 𝑂𝑂𝑂𝑂 and 𝐺𝐺𝐺𝐺 of 𝐴𝐴𝐵𝐵 and 𝐶𝐶𝐶𝐶
5)
6) 6) Neither 𝐴𝐴𝐵𝐵 𝑎𝑎𝑠𝑠𝑎𝑎 𝐶𝐶𝐶𝐶 , nor 𝑂𝑂𝑂𝑂𝑎𝑎𝑠𝑠𝑎𝑎 𝐺𝐺𝐺𝐺 are parallel
G
O
D
E
F
6 in
4 in A B C
2in
O
D C
B
A
G
E
F
H
-
34) Given: ABCD is a kite inscribed in a circle center O Prove: ∠ A and ∠C are right angles
STATEMENTS REASONS
1) ABCD is a kite inscribed in a circle O
1) Given
2) 𝐴𝐴𝐵𝐵 ≅ 𝐵𝐵𝐶𝐶; 𝐴𝐴𝐶𝐶 ≅ 𝐶𝐶𝐶𝐶 2)
3) 3) Construction
4) 𝑂𝑂𝐴𝐴 ≅ 𝑂𝑂𝐵𝐵 ≅ 𝑂𝑂𝐶𝐶 ≅ 𝑂𝑂𝐶𝐶 4)
5) ∆OAB ≅ ∆OBC; ∆OAD ≅ ∆OCD
5)
6) 6) CPCTC
7) m∠ 1 = m∠2; m∠ 3 = m∠4 7)
8) m∠ 1 + m∠2 + m∠ 3 + m∠4 = 3600 8) Angles form a revolution
9) 2 m∠ 1 + 2 m∠4= 3600 9)
10) 10) Divide by 2
11) 𝐴𝐴𝐶𝐶 is a diameter 11) By 10)
12) 12) An angle inscribed in a semi-circle is a right angle
A
B
C
D
O
A
B
C
D
1 2
4 3 O
-
35) In the Great Stone Bridge (also called the Zhaozhou Bridge) in southern Hebei province, China, built by Li Chun between 589 and 618, the single span of 123 feet has a rise of only 23 feet from the abutments to the crown. Find the radius of the curvature of the bridge.
36) The bridge has dimensions as shown. Find the radius of the curvature of the bridge.
37) Two concentric circles and the square whose sides are tangent to the smaller circle. If the radius of the large circle is 1 foot, find the radius of the small circle. 38) Two concentric circles and the square whose sides are tangent to the smaller circle. If the radius of the small circle is 1 foot, find the radius of the large circle. Exercise 35 and 36
123ft
23ft
r
D
61.5ft
23ft
O
A B C
https://www.britannica.com/topic/Great-Stone-Bridgehttps://www.britannica.com/place/Hebei
-
39) Prove that The measure of a special inscribed angle, an angle formed by a chord and a tangent intersecting at the tangent point, is one-half the measure of its intercepted arc.
Hint: Draw the diameter AD 40) A person in an orbiting spacecraft (see figure) h = 616 mi above the earth sights the horizon on the earth at an angle of depression of α = 300. Find the radius of the earth.
SUMMARY 41) If a chord is ______________ to the radius, the radius _______the chord. 42) Two tangents to a circle from an _________point are ____________.
43) If a radius ___________the chord, the radius is ___________to the chord.
44) The ____________ is perpendicular to the radius at the _______ ______.
O
A
B
C
-
6.3 More Angles of circles
Outline: Theorems
Construction
1) Theorems
a) Theorem 1
The measure of an angle formed by two chords intersecting within a circle is equal to one half the sums of the measures of the intercepted arcs.
Proof
Given: AB and CD chords
Prove: BD)mAC(m211 m ∠+∠=∠
STATEMENTS REASONS
1) AB and CD chords 1) Given
2) Draw CB 2) Construction
3) 3m2m1 m ∠+∠=∠ 3) Measure of ext. ∠ of a Δ = sum of measures of two non- adjacent s'∠
4) ACm213m ∠=∠ ; BDm
212m ∠=∠
3) The measure of an inscribed angle of a circle is one half the measure of its intercepted arc.
4) ACm211 m ∠=∠ BDm
21
∠+ 4) Substitution 4) into 3)
5) BD)mAC(m211 m ∠+∠=∠
5) Simplify
A
B
D
C 2
3
1 E
-
b) Theorem 2
The measure of an angle formed by two secants intersecting at a point outside the circle is equal to one half the differences of the measures of its intercepted arcs.
The proof is left as exercise.
c) Theorem 3
The measure of an angle formed by a secant and a tangent intersecting at a point outside the circle is equal to one half the differences of the measures of its intercepted arcs.
The proof is left as exercise
d) Theorem 4
If two chords intersect within a circle, then the product of the lengths of the parts of one chord is equal to the product of those of the other chord.
Given: AB and CD chords intersecting at E
Prove: (AE)( EB )= (EC)(ED)
STATEMENTS REASONS
1) AB and CD chords intersecting at E 1) Given
2) Draw AC and BD 2) Construction
3) DA ∠≅∠ ; BC ∠≅∠ 3) 2 inscribed s'∠ intercept same arc, they’re ≅
4) DBE ΔACE ∆ 4) AA
5) EBEC
EDAE
= 5)CPCTC
6) AE EB = EC ED 6) Cross multiply
D
A
B
C
E
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Example: Archimedes’ trisected angle.
Suppose that we want to trisect an angle A∠ . From A draw a semi-circle that intersects two sides of the angle A∠ at B and C. Extend the side BA to cut the semi-circle at F. Now from C draw a secant to intersects the circle at D and the side AB at E such that AD = ED. We will prove that CEB3mCABm ∠=∠
Given: A center of a semi-circle; ED = AD
Prove: CEB3mCABm ∠=∠
STATEMENTS REASONS
1) A center of a semi-circle; ED = AD 1) Given
2) ED ≅ AD 2) Definition of congruent segments
3) CEBDAE ∠≅∠ 3) In a triangle, the opp. s'∠ to ≅ sides are ≅
4) CEBmDAEm ∠=∠ 4) Definition of congruent angles
5) BCmCABm ∠=∠ ; FDmDAFm ∠=∠
5) Measure of central∠ of a circle = measures of intercepted arc
6) BCm21CEB m ∠=∠ FDm
21
∠− 6) The measure of an angle formed by two secants intersecting outside a circle is equal to one half the differences of the measures of the intercepted arcs.
7) CABm21CEB m ∠=∠ DAFm
21
∠− 7) Substitution 5) into 6)
8) CABmCEB 2m ∠=∠ DAFm∠− 8) Multiply by 2 CB s'∠ Δ
9) CABmCEB 2m ∠=∠ CEBm∠− 9) Substitution 4) into 7)
10) CABmCEB 3m ∠=∠ 10) Simplify
r r r
E A
D
C
B F
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d) Theorem 5
If two chords intersect outside a circle, then the product of the lengths of the parts of one chord is equal to the product of those of the other chord.
Given: AC and AE chords intersecting at A
Prove: (AB)( AC) =( AE)( AD)
STATEMENTS REASONS
1) AC and AE chords intersecting at A 1) Given
2) Draw DC and BE 2) Construction
3) EC ∠≅∠ 3) 2 inscribed s'∠ intercept same arc, they’re ≅
4) AA ∠≅∠ 4) Identity
5) ∆ACD ∾ ∆AEB 5) AA
6) ACAD
AEAB
= 6)CPCTC
7) (AB)( AC) =( AE) (AD) 7) Cross multiply
A
B
C
D E
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Example 6: The chords AB and CD intersect at point E, DQ , and FG are tangents to the circle. If the
measures of arcs 00 58DB ,92AC ==∩∩
,FD = 4, AD = 5 and 040=∠ADQm . Find
a. ADCm∠
b. AFCm∠
c. AECm∠
d.∩
BCm
e. FG.
This example is left as an exercise.
Example 7: A proof of Law of Cosine by Math Professor Roger Nelson.
Given: A circle center at O with radius b, OF = c, FB = a, ∠ABC = Ɵ
Prove: 𝑠𝑠2 = 𝑎𝑎2 + 𝑏𝑏2 − 2𝑎𝑎𝑏𝑏𝑠𝑠𝑠𝑠𝑠𝑠𝑎𝑎 Arithmetic Proof
Let CF = x. Since ∆ABC is a right triangle (Why ?), we have
𝑠𝑠𝑠𝑠𝑠𝑠𝑎𝑎 = 𝐴𝐴𝐶𝐶𝐴𝐴𝐴𝐴
= 𝑎𝑎+𝑥𝑥2𝑏𝑏
⇒ 2𝑠𝑠𝑠𝑠𝑠𝑠𝑎𝑎 = 𝑎𝑎𝑏𝑏
+ 𝑥𝑥𝑏𝑏
(*)
By theorem 3, we get
(FD)(FE) = (BF)(FC) or (b – c )(b + c ) = ax
It follows that
𝑏𝑏2−𝑐𝑐2
𝑎𝑎= 𝑥𝑥 (**)
Substituting equation (**) into equation (*), we obtain
2𝑠𝑠𝑠𝑠𝑠𝑠𝑎𝑎 =𝑎𝑎𝑏𝑏
+𝑏𝑏2 − 𝑠𝑠2
𝑎𝑎𝑏𝑏=
𝑎𝑎2 + 𝑏𝑏2 − 𝑠𝑠2
𝑎𝑎𝑏𝑏
Hence 𝑠𝑠2 = 𝑎𝑎2 + 𝑏𝑏2 − 2𝑎𝑎𝑏𝑏𝑠𝑠𝑠𝑠𝑠𝑠𝑎𝑎
G
Q
O b
c a
A
F
E
D C
B Ɵ
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Example 8: The billboard of height b is mounted on the side of the building with bottom edge at a distance h from the street as in the figure. At what distance x should observer stand in front of the wall to maximize the angle of observation. Hint prove θ ≅ ψ
In triangle APB, ABPmθmψm ∠+∠=∠ . Why?
Or ABPmψmθm ∠−∠=∠ . Thus, max of θ occurs if θmψm ∠=∠ . Why?
We also have: ( )xhBRQtan =∠ and ( )
bhxRCQtan+
=∠ . But BRQRCQ ∠≅∠ . Why?
It follows ⇒=+ x
hbh
x )( bhhx += .
To get the best view of the Statue of Liberty with b = 92 m, h = 46 m, the distance from the base
67.79)4692(46)( ≈+=+= bhhx m.
Also, =∠θm =
−
−−
xBQ
xCQ 11 tantan ≈
−
+ −−
67.7946tan
67.794692tan 11 300.
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2) Construction
a) Tangent at a point P on the circle center O.
Draw the radius OP and extend it to form a ray→
OX .
Use P as the center, we draw two arcs with radius less than OP. Two arcs cut the ray →
OX at Q and S.
Now we construct ⊥→
WP→
OX at P. →
WP is the tangent line to the circle at P
b) Tangent line from the point P outside of the circle center at O.
Draw OP . From O and P, we draw two arcs with radius less than OP and they cut each other at Q and S. Connect Q and S, we have the midpoint M of the segment OP
From M, we draw a circle with radius MO. The circle cuts the given circle at T and W. Connect PT and PW, we have two tangents.
P
O
X W
Q
S
P O
W Q
S
M
T
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HOMEWORK 6.3
Refer to the following figure for Exercise 1 – 4
1) If 𝑀𝑀𝑃𝑃� = 380 and 𝑀𝑀𝑀𝑀� = 850, find m ∠ P.
2) If 𝑀𝑀𝑃𝑃� = 250 and m ∠ P = 300, find 𝑀𝑀𝑀𝑀� .
3) If PN = 14”, PM = 6”, and PQ = 8”, find PR.
4) If MN = 4”, PM = 6”, and PQ = 8”, find QR.
Refer to the following figure for Exercise 5 – 9
5) If 𝑀𝑀𝑃𝑃� = 350 and 𝑀𝑀𝑀𝑀� = 600, find m ∠ 1.
6) If 𝑀𝑀𝑃𝑃� = 250 and m ∠ 1 = 300, find 𝑀𝑀𝑀𝑀� .
7) If QS = 7 cm, SM = 6 cm, and RS = 15 cm, find NS.
9) If MR = 16 cm, SM = 3 cm, and NS = 12 cm, find SQ.
Refer to the following figure for Exercise 9 – 13. Point O is the center of the circle.
10) If ∠1 ≅ ∠ONM and𝑀𝑀𝑃𝑃 bisects ∠ONM, find m∠ P and m 𝑀𝑀𝑃𝑃�
11) If m∠P = 340, find m ∠ NQP
12) If QR = 10”, QP = 6” and PM = 7”, find MN.
13) If QR = 12”, RP = 20” and PN = 18”, find MN.
M
P
N
Q
R
M
P
N
Q R 1 S
M
P
N
Q
R O
1
2 3
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The chords AB and CD intersect at point E, DQ , and FG are tangents to the circle. If m 𝐴𝐴𝐶𝐶� = 920 , 𝑚𝑚 𝐶𝐶𝐵𝐵� = 580, FD = 4 cm, AD = 5 cm and 040=∠ADQm . Find
14) ADCm∠
15) AFCm∠
16) AECm∠
17) 𝑚𝑚 𝐵𝐵𝐶𝐶�
18) FG.
19) The measure of an angle formed by two secants intersecting at a point outside the circle is equal to one half the differences of the measures of its intercepted arcs.
Given: AB and CD are secants intersecting at E
Prove: 𝑚𝑚∠𝑂𝑂 = 12
�𝑚𝑚𝐵𝐵𝐶𝐶� − 𝑚𝑚𝐴𝐴𝐶𝐶� �
STATEMENTS REASONS
1) 1)
2) Draw𝐶𝐶𝐵𝐵 2)
3) 𝑚𝑚∠1 = 𝑚𝑚∠2 + 𝑚𝑚 ∠𝑂𝑂 3)
4) 4) Subtraction
5) 𝑚𝑚∠1 = 12
𝑚𝑚𝐵𝐵𝐶𝐶� ; 𝑚𝑚∠2 = 12
𝑚𝑚𝐴𝐴𝐶𝐶� 5)
6) 6) Substitution 5) into 4)
7) 7)
G
Q
E
A
B
D C
1
2
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20) Given: BC = c, BQ = h, and PQ = x.
Prove: )( bhhx +=
STATEMENTS REASONS
1) BC = c, BQ = h, and PQ = x 1) Given
2) Draw a circle that has a chord 𝐵𝐵𝐶𝐶 and tangent 𝑁𝑁𝑃𝑃;
2)
3) ∠CAB ≅ ∠ CRB 3)
4) m ∠ CAB = m ∠ Ɵ + m ∠ABP 4)
5) m ∠ Ɵ = m ∠CAB − m ∠ ABP 5)
6)m ∠Ɵ = m ∠ CAB 6) If max Ɵ occurs, m ∠ ABP = 0
7) ∠ CAB ≅ ∠ Ɵ 7)
8) P → 𝑀𝑀 or PQ ≡ 𝑀𝑀𝑃𝑃 = 𝑥𝑥 8) By 6)
9) ( )xhBRQtan =∠ ; ( )
bhxRCQtan+
=∠ 9)
10) BRQRCQ ∠≅∠ 10)
11) ℎ𝑥𝑥
= 𝑥𝑥ℎ+𝑏𝑏
11)
12) 12) Cross multiply
13) 13)
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21) If you want to get the best view of the Statue of Liberty with b = 92 m, h = 46 m, what is the distance from the base?
22) If you want to get the best view of the billboard with b = 14ft, h = 46ft, what is the distance from the base?
23) Prove that the measure of an angle formed by a secant and a tangent intersecting at a point outside the circle is equal to one half the differences of the measures of its intercepted arcs.
24) Prove that if a parallelogram is inscribed in a circle, it’s a rectangle.
25) Given: 𝐴𝐴𝐵𝐵 is common tangent, 𝑂𝑂𝑃𝑃 is a line of centers; 𝐴𝐴𝐵𝐵𝑎𝑎𝑠𝑠𝑎𝑎𝑂𝑂𝑃𝑃𝑠𝑠𝑠𝑠𝑡𝑡𝑖𝑖𝑖𝑖𝑠𝑠𝑖𝑖𝑠𝑠𝑡𝑡 𝑎𝑎𝑡𝑡 𝐶𝐶
Prove: ∆OAC ∆QBC
26) Given: 𝐴𝐴𝐵𝐵 ≅ 𝐶𝐶𝐶𝐶 in the circle with center O. Prove: ∆𝐴𝐴𝐵𝐵𝐶𝐶 ≅ ∆𝐶𝐶𝐶𝐶𝐵𝐵 27) Given: Tangents 𝐴𝐴𝐵𝐵, 𝐵𝐵𝐶𝐶, and 𝐴𝐴𝐶𝐶 to the circle with center o at points M, N, ad P respectively AB = 14, BC = 16, and AC = 12. Find: AM, PC, and BN Exercise 26 Exercise 27
A
B C
M P
N
A
B
O Q
C
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28) Given: 𝐴𝐴𝐵𝐵 and 𝑀𝑀𝐶𝐶 are an external tangent to circles center O and Q (see figure) Prove: 𝑀𝑀