This document provides instruction on using the Law of Sines to solve triangles. It begins with examples of using the Law of Sines to find missing side lengths or angle measures when two angles and a side, or two sides and an angle are known. It also covers cases where an ambiguous triangle could result from given side-side-angle information. The document demonstrates solving for the area of triangles using trigonometric functions. It concludes with practice problems applying the Law of Sines to find missing measurements and the number of possible triangles based on given side lengths and an angle measure.

1. Holt Algebra 2
UNIT 10.4 LAW OF SINESUNIT 10.4 LAW OF SINES

2. 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, m∠A = 122° and m∠B =17°. What is the
m∠C ?
5.
28.5 10
41°

3. 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

4. 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.

5. 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.

6. 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

8. 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
.

9. Check It Out! Example 1
Find the area of the triangle.
Round to the nearest tenth.
Area = ac sin B
≈ 47.88307441
Write the area formula.
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
.

10. 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.

12. 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

13. 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.
m∠D + m∠E + m∠F = 180°
33° + m∠E + 28° = 180°
m∠E = 119°
Triangle Sum Theorem.
Substitute 33° for m∠D
and 28° for m∠F.
Solve for m∠E.

14. Example 2A Continued
Step 2 Find the unknown side lengths.
sin D sin F
d 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 ≈ 27.9
Solve for the
unknown side.
Law of
Sines.
Substitute.
Cross
multiply.

15. 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.
m∠P = 180° – 36° – 39° = 105° Triangle Sum
Theorem
Q
r

16. 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 Q
p 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

17. Check It Out! Example 2a
Solve the triangle. Round
to the nearest tenth.
Step 1 Find the third angle measure.
m∠K = 31° Solve for m∠K.
m∠H + m∠J + m∠K = 180°
42° + 107° + m∠K = 180°
Substitute 42° for m∠H
and 107° for m∠J.

18. Check It Out! Example 2a Continued
Step 2 Find the unknown side lengths.
sin H sin J
h 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 ≈ 6.5
Solve for the
unknown side.
Law of
Sines.
Substitute.
Cross
multiply.

19. Check It Out! Example 2b
Solve the triangle. Round
to the nearest tenth.
Step 1 Find the third angle
measure.
m∠N = 180° – 56° – 106° = 18° Triangle Sum
Theorem

20. Check It Out! Example 2b
Solve the triangle. Round
to the nearest tenth.
Step 2 Find the unknown side lengths.
sin N sin M
n 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=

21. 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.

23. Solving a Triangle Given a, b, and m∠A

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

25. Example 3: Art Application
Determine the number of triangular banners
that can be formed using the measurements a
= 50, b = 20, and m∠A = 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

26. Example 3 Continued
Step 2 Determine m∠B.
Law of Sines
Substitute.
Solve for sin B.

27. 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 m∠B.
Step 3 Find the other unknown measures of the
triangle.
Solve for m∠C.
28° + 10.8° + m∠C = 180°
m∠C = 141.2°
m B = Sin-1

28. Example 3 Continued
Solve for c.
c ≈ 66.8
Law of Sines
Substitute.
Solve for c.

29. Check It Out! Example 3
Determine the number of triangles Maggie can
form using the measurements a = 10 cm, b =
6 cm, and m∠A =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.

30. Check It Out! Example 3 Continued
Step 2 Determine m∠B.
Law of Sines
Substitute.
Solve for sin B.
sin B ≈ 0.58

31. Check It Out! Example 3 Continued
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 m∠C.
105° + 35.4° + m∠C = 180°
m∠C = 39.6°

32. Check It Out! Example 3 Continued
Solve for c.
c ≈ 6.6 cm
Law of Sines
Substitute.
Solve for c.

33. 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;
m∠C = 133.8°

34. 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
m∠A = 39°. Solve each triangle. Round to
the nearest tenth.
2;
c1 ≈ 21.7 cm;
m∠B1 ≈ 64.0°;
m∠C1 ≈ 77.0°;
c2 ≈ 9.4 cm;
m∠B2 ≈ 116.0°;
m∠C2 ≈ 25.0°

35. All rights belong to their
respective owners.
Copyright Disclaimer Under
Section 107 of the
Copyright Act 1976,
allowance is made for "fair
use" for purposes such as
criticism, comment, news
reporting, TEACHING,
scholarship, and research.
Fair use is a use permitted
by copyright statute that
might otherwise be
infringing.
Non-profit, EDUCATIONAL
or personal use tips the
balance in favor of fair use.