Holt Algebra 2

13-5 The Law of Sines13-5 The Law of Sines

Holt Algebra 2

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Holt Algebra 2

13-5 The Law of Sines

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

Holt Algebra 2

13-5 The Law of Sines

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.

Holt Algebra 2

13-5 The Law of Sines

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.

Holt Algebra 2

13-5 The Law of Sines

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

Holt Algebra 2

13-5 The Law of Sines

Holt Algebra 2

13-5 The Law of Sines

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

Write the area formula.

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.

Holt Algebra 2

13-5 The Law of Sines

Check It Out! Example 1

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

Holt Algebra 2

13-5 The Law of Sines

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.

Holt Algebra 2

13-5 The Law of Sines

Holt Algebra 2

13-5 The Law of Sines

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

Holt Algebra 2

13-5 The Law of Sines

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.

Holt Algebra 2

13-5 The Law of Sines

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.

Holt Algebra 2

13-5 The Law of Sines

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

Solve the triangle. Round to the nearest tenth.

Step 1 Find the third angle measure.

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

Q

r

Holt Algebra 2

13-5 The Law of Sines

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

Solve the triangle. Round to the nearest tenth.

Step 2 Find the unknown side lengths.

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

Holt Algebra 2

13-5 The Law of Sines

Check It Out! Example 2a

Solve the triangle. Round to the nearest tenth.

Holt Algebra 2

13-5 The Law of Sines

Check It Out! Example 2b

Solve the triangle. Round to the nearest tenth.

Holt Algebra 2

13-5 The Law of Sines

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.

Holt Algebra 2

13-5 The Law of Sines

Holt Algebra 2

13-5 The Law of Sines

Solving a Triangle Given a, b, and mA

Holt Algebra 2

13-5 The Law of Sines

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

Remember!

Holt Algebra 2

13-5 The Law of Sines

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

Holt Algebra 2

13-5 The Law of Sines

Example 3 Continued

Step 2 Determine mB.

Law of Sines

Substitute.

Solve for sin B.

Holt Algebra 2

13-5 The Law of Sines

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

Holt Algebra 2

13-5 The Law of Sines

Example 3 Continued

Solve for c.

c ≈ 66.8

Law of Sines

Substitute.

Solve for c.

Holt Algebra 2

13-5 The Law of Sines

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.