holt mcdougal algebra 2 10-5 the law of sines 10-5 the law of sines holt algebra 2 warm up warm up...

Holt McDougal Algebra 2

10-5 The Law of Sines10-5 The Law of Sines

Holt Algebra 2

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10-5 The Law of Sines

Warm Up

Find the area of each triangle with the given base and height.

1. b =10, h = 7 2. b = 8, h = 4.6

35 units2 18.4 units2

Solve each proportion.

3. 4.

In ∆ABC, mA = 122° and mB =17°. What is the mC ?

5.

28.5 10

41°



Determine the area of a triangle given side-angle-side information.

Use the Law of Sines to find the side lengths and angle measures of a triangle.

Objectives



The area of the triangle representing the sail is Although you do not know the value of h, you can calculate it by using the fact that sin A = , or h = c sin A.

A sailmaker is designing a sail that will have the dimensions shown in the diagram. Based on these dimensions, the sailmaker can determine the amount of fabric needed.



Area =

Area =

This formula allows you to determine the area of a triangle if you know the lengths of two of its sides and the measure of the angle between them.

Write the area formula.

Substitute c sin A for h.



An angle and the side opposite that angle are labeled with the same letter. Capital letters are used for angles, and lowercase letters are used for sides.

Helpful Hint



Example 1: Determining the Area of a Triangle

Find the area of the triangle. Round to the nearest tenth.

Area = ab sin C

≈ 4.820907073


Substitute 3 for a, 5 for b, and 40° for C.

Use a calculator to evaluate the expression.

The area of the triangle is about 4.8 m2.



Check It Out! Example 1

Find the area of the triangle. Round to the nearest tenth.

Area = ac sin B

≈ 47.88307441


Substitute 8 for a, 12 for c, and 86° for B.

Use a calculator to evaluate the expression.

The area of the triangle is about 47.9 m2.



The area of ∆ABC is equal to bc sin A or ac sin B or ab sin C. By setting these expressions equal to each other, you can derive the Law of Sines.

bc sin A = ac sin B = ab sin C

bc sin A = ac sin B = ab sin C

bc sin A ac sin B ab sin C abc abc abc

= =

sin A = sin B = sin C a b c

Multiply each expression by 2.

Divide each expression by abc.

Divide out common factors.



The Law of Sines allows you to solve a triangle as long as you know either of the following:

1. Two angle measures and any side length–angle-angle-side (AAS) or angle-side-angle (ASA) information

2. Two side lengths and the measure of an angle that is not between them–side-side-angle (SSA) information



Example 2A: Using the Law of Sines for AAS and ASA

Solve the triangle. Round to the nearest tenth.

Step 1. Find the third angle measure.

mD + mE + mF = 180°

33° + mE + 28° = 180°

mE = 119°

Triangle Sum Theorem.

Substitute 33° for mD and 28° for mF.

Solve for mE.



Example 2A Continued

Step 2 Find the unknown side lengths.

sin D sin Fd f

=sin E sin F

e f=

sin 33° sin 28°d 15=

sin 119° sin 28°e 15=

d sin 28° = 15 sin 33° e sin 28° = 15 sin 119°

d = 15 sin 33°sin 28°

d ≈ 17.4

e = 15 sin 119°sin 28°

e ≈ 27.9Solve for the

unknown side.

Law of Sines.

Substitute.

Crossmultiply.



Example 2B: Using the Law of Sines for AAS and ASA


Step 1 Find the third angle measure.

mP = 180° – 36° – 39° = 105° Triangle Sum Theorem

Q

r



Example 2B: Using the Law of Sines for AAS and ASA



sin P sin Qp q= sin P sin R

p r=Law of Sines.

sin 105° sin 36°10 q= sin 105° sin 39°

10 r=Substitute.

q = 10 sin 36°sin 105°

≈ 6.1 r = 10 sin 39°sin 105°

≈ 6.5

Q

r



Check It Out! Example 2a



mK = 31° Solve for mK.

mH + mJ + mK = 180°

42° + 107° + mK = 180°Substitute 42° for mH

and 107° for mJ.



Check It Out! Example 2a Continued


sin H sin Jh j

=sin K sin H

k h=

sin 42° sin 107°h 12=

sin 31° sin 42°k 8.4=

h sin 107° = 12 sin 42° 8.4 sin 31° = k sin 42°

h = 12 sin 42°sin 107°

h ≈ 8.4

k = 8.4 sin 31°sin 42°

k ≈ 6.5Solve for the

unknown side.

Law of Sines.

Substitute.

Crossmultiply.



Check It Out! Example 2b



mN = 180° – 56° – 106° = 18° Triangle Sum Theorem



Check It Out! Example 2b



sin N sin Mn m= sin M sin P

m p=Law of Sines.

Substitute.sin 18° sin 106°1.5 m=

m = 1.5 sin 106°sin 18°

≈ 4.7 p = 4.7 sin 56°sin 106°

≈ 4.0

sin 106° sin 56°4.7 p=



When you use the Law of Sines to solve a triangle for which you know side-side-angle (SSA) information, zero, one, or two triangles may be possible. For this reason, SSA is called the ambiguous case.



Solving a Triangle Given a, b, and mA



When one angle in a triangle is obtuse, the measures of the other two angles must be acute.

Remember!



Example 3: Art Application

Determine the number of triangular banners that can be formed using the measurements a = 50, b = 20, and mA = 28°. Then solve the triangles. Round to the nearest tenth.

Step 1 Determine the number of possible triangles. In this case, A is acute.

Because b < a; only one triangle is possible.

A B

C

b a

c



Example 3 Continued

Step 2 Determine mB.

Law of Sines

Substitute.

Solve for sin B.



Example 3 Continued

Let B represent the acute angle with a sine of 0.188. Use the inverse sine function on your calculator to determine mB.

Step 3 Find the other unknown measures of the triangle.

Solve for mC.

28° + 10.8° + mC = 180°

mC = 141.2°

m B = Sin-1



Example 3 Continued

Solve for c.

c ≈ 66.8

Law of Sines

Substitute.

Solve for c.



Check It Out! Example 3

Determine the number of triangles Maggie can form using the measurements a = 10 cm, b = 6 cm, and mA =105°. Then solve the triangles. Round to the nearest tenth.

Step 1 Determine the number of possible triangles. In this case, A is obtuse.

Because b < a; only one triangle is possible.



Check It Out! Example 3 Continued

Step 2 Determine mB.

Law of Sines

Substitute.

Solve for sin B.

sin B ≈ 0.58




m B = Sin-1

Let B represent the acute angle with a sine of 0.58. Use the inverse sine function on your calculator to determine m B.

Step 3 Find the other unknown measures of the triangle.

Solve for mC.

105° + 35.4° + mC = 180°

mC = 39.6°




Solve for c.

c ≈ 6.6 cm

Law of Sines

Substitute.

Solve for c.



Lesson Quiz: Part I

1. Find the area of the triangle. Round to the nearest tenth.

17.8 ft2

2. Solve the triangle. Round to the nearest tenth.

a 32.2; b 22.0; mC = 133.8°



Lesson Quiz: Part II

3. Determine the number of triangular quilt pieces that can be formed by using the measurements a = 14 cm, b = 20 cm, and mA = 39°. Solve each triangle. Round to the nearest tenth. 2;

c1 21.7 cm;mB1 ≈ 64.0°;mC1 ≈ 77.0°;

c2 ≈ 9.4 cm;mB2 ≈ 116.0°;mC2 ≈ 25.0°

holt mcdougal algebra 2 10-5 the law of sines 10-5 the law of sines holt algebra 2 warm up warm up...

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