10-5The
10-5
TheLaw
LawofofSines
Sines
Warm Up
Lesson Presentation
Lesson Quiz
HoltMcDougal
Algebra 2Algebra 2
Holt
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.
28.5
4.
10
5. In ∆ABC, mA = 122° and mB =17°. What is the
mC ? 41°
Holt McDougal Algebra 2
10-5 The Law of Sines
Objectives
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.
Holt McDougal Algebra 2
10-5 The Law of Sines
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.
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.
Holt McDougal Algebra 2
10-5 The Law of Sines
Area =
Area =
Write the area formula.
Substitute c sin A for h.
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.
Holt McDougal Algebra 2
10-5 The Law of Sines
Helpful Hint
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.
Holt McDougal Algebra 2
10-5 The Law of Sines
Holt McDougal Algebra 2
10-5 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
Write the area formula.
Substitute 3 for a, 5 for b,
and 40° for C.
≈ 4.820907073
Use a calculator to evaluate
the expression.
The area of the triangle is about 4.8 m2.
Holt McDougal Algebra 2
10-5 The Law of Sines
Check It Out! Example 1
Find the area of the triangle.
Round to the nearest tenth.
Area =
ac sin B
Write the area formula.
Substitute 8 for a, 12 for c,
and 86° for B.
≈ 47.88307441
Use a calculator to evaluate
the expression.
The area of the triangle is about 47.9 m2.
Holt McDougal Algebra 2
10-5 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
b
c
a
Holt McDougal Algebra 2
Multiply each
expression by 2.
Divide each
expression by abc.
Divide out common
factors.
10-5 The Law of Sines
Holt McDougal Algebra 2
10-5 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–angleangle-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 McDougal Algebra 2
10-5 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.
mD + mE + mF = 180°
Triangle Sum Theorem.
33° + mE + 28° = 180°
Substitute 33° for mD
and 28° for mF.
Solve for mE.
mE = 119°
Holt McDougal Algebra 2
10-5 The Law of Sines
Example 2A Continued
Step 2 Find the unknown side lengths.
sin F
sin D
d =
f
sin 33° sin 28°
d =
15
Law of
Sines.
Substitute.
sin F
sin E
e =
f
sin 119° sin 28°
e =
15
Cross e sin 28° = 15 sin 119°
multiply.
15 sin 119°
15 sin 33°
e
=
d=
sin 28°
sin 28°
Solve for the
d ≈ 17.4
e ≈ 27.9
unknown side.
d sin 28° = 15 sin 33°
Holt McDougal Algebra 2
10-5 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.
mP = 180° – 36° – 39° = 105°
Holt McDougal Algebra 2
r
Q
Triangle Sum
Theorem
10-5 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 Q
p
q
Law of
Sines.
r
Q
sin P = sin R
p
r
sin 105° = sin 39°
sin 105° = sin 36°
Substitute.
r
q
10
10
10 sin 36°
10 sin 39°
q=
r=
≈ 6.1
≈ 6.5
sin 105°
sin 105°
Holt McDougal Algebra 2
10-5 The Law of Sines
Check It Out! Example 2a
Solve the triangle. Round
to the nearest tenth.
Step 1 Find the third angle measure.
mH + mJ + mK = 180°
42° + 107° + mK = 180°
mK = 31°
Holt McDougal Algebra 2
Substitute 42° for mH
and 107° for mJ.
Solve for mK.
10-5 The Law of Sines
Check It Out! Example 2a Continued
Step 2 Find the unknown side lengths.
sin J
sin H
h =
j
sin 42° sin 107°
h =
12
Law of
Sines.
Substitute.
sin H
sin K
k =
h
sin 42°
sin 31°
k =
8.4
Cross 8.4 sin 31° = k sin 42°
h sin 107° = 12 sin 42° multiply.
8.4 sin 31°
12 sin 42°
k
=
h=
sin 42°
sin 107°
Solve for the
h ≈ 8.4
k ≈ 6.5
unknown side.
Holt McDougal Algebra 2
10-5 The Law of Sines
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°
Holt McDougal Algebra 2
Triangle Sum
Theorem
10-5 The Law of Sines
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
Law of
Sines.
sin M= sin P
m
p
sin 106° = sin 56°
sin 18° = sin 106°
Substitute.
p
m
4.7
1.5
1.5 sin 106°
4.7 sin 56°
m=
p
=
≈ 4.7
≈ 4.0
sin 18°
sin 106°
Holt McDougal Algebra 2
10-5 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 McDougal Algebra 2
10-5 The Law of Sines
Holt McDougal Algebra 2
10-5 The Law of Sines
Solving a Triangle Given a, b, and mA
Holt McDougal Algebra 2
10-5 The Law of Sines
Remember!
When one angle in a triangle is obtuse, the
measures of the other two angles must be acute.
Holt McDougal Algebra 2
10-5 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 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.
C
b
A
a
c
B
Because b < a; only one triangle is possible.
Holt McDougal Algebra 2
10-5 The Law of Sines
Example 3 Continued
Step 2 Determine mB.
Law of Sines
Substitute.
Solve for sin B.
Holt McDougal Algebra 2
10-5 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 mB.
m
B = Sin-1
Step 3 Find the other unknown measures of
the triangle.
Solve for mC.
28° + 10.8° + mC = 180°
mC = 141.2°
Holt McDougal Algebra 2
10-5 The Law of Sines
Example 3 Continued
Solve for c.
Law of Sines
Substitute.
Solve for c.
c ≈ 66.8
Holt McDougal Algebra 2
10-5 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 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.
Holt McDougal Algebra 2
10-5 The Law of Sines
Check It Out! Example 3 Continued
Step 2 Determine mB.
Law of Sines
Substitute.
Solve for sin B.
sin B ≈ 0.58
Holt McDougal Algebra 2
10-5 The Law of Sines
Check It Out! Example 3 Continued
Let B represent the acute angle with a sine of
0.58. Use the inverse sine function on your
calculator to determine m B.
m
B = Sin-1
Step 3 Find the other unknown measures of
the triangle.
Solve for mC.
105° + 35.4° + mC = 180°
mC = 39.6°
Holt McDougal Algebra 2
10-5 The Law of Sines
Check It Out! Example 3 Continued
Solve for c.
Law of Sines
Substitute.
Solve for c.
c ≈ 6.6 cm
Holt McDougal Algebra 2
10-5 The Law of Sines
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°
Holt McDougal Algebra 2
10-5 The Law of Sines
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°
Holt McDougal Algebra 2