9.5 Apply the Law of Sines
When can the law of sines be used to
solve a triangle?
How is the SSA case different from the
AAS and ASA cases?
Law of Sines
The law of sines can be used to solve triangles when two
angles and the length of any side are known. (AAS or ASA
cases), or when the lengths of two sides and an angle opposite
one of the two side are known (SSA case).
Solve
ABC with C = 107°, B = 25°, and b = 15.
SOLUTION
First find the angle: A = 180° – 107° – 25° = 48°.
By the law of sines, you can write
a
15
c
=
=
sin 48°
sin 25° sin 107°
a
15 Write two equations,
=
sin 48°
sin 25° each with one variable.
c
15
=
sin 107°
sin 25°
15 sin 48°
15 sin 107°
Solve for each variable. c =
a =
sin 25°
sin 25°
a
26.4
ANSWER In
Use a calculator.
ABC, A = 48°, a
c
33.9
26.4, and c
33.9.
Solve ABC.
SOLUTION
1.
B = 34°, C = 100°, b = 8
First find the angle: A = 180° – 34° – 100° = 46°.
By the law of sines, you can write
a
8
c
=
=
sin 46°
sin 34° sin 100°
a
8
c
8
Write two equations, each
=
=
sin 46°
sin 34° with one variable.
sin 100°
sin 34°
8 sin 46°
a =
sin 34°
a
8 sin 100°
Solve for each variable. c =
sin 34°
10.3
c
14.1
Use a calculator.
ANSWER
In
ABC, A 46°, a
10.3, and c
14.1.
Solve
ABC.
2.
A = 51°, B = 44°, c = 11
SOLUTION
First find the angle: C = 180° – 51° – 44° = 85°.
By the law of sines, you can write
a
b
11
=
=
sin 51°
sin 44°
sin 85°
a
11 Write two equations, each
b
11
=
=
sin 51°
sin 85° with one variable.
sin 44°
sin 85°
11 sin 51°
11 sin 44°
a =
sin 85° Solve for each variable. b =
sin 85°
a
8.6
ANSWER In
Use a calculator.
ABC, A 85°, a
b
7.7
8.6, and b
7.7.
SSA Case
Two angles and one side (AAS or ASA) determine exactly one triangle.
Two sides and an angle opposite one of the sides (SSA) may determine:
• no triangle
• one triangle
• two triangles.
Solve
ABC with A = 115°, a = 20, and b = 11.
SOLUTION
First make a sketch. Because A is obtuse and the side
opposite A is longer than the given adjacent side, you
know that only one triangle can be formed. Use the law of
sines to find B.
sin B
sin 115°
=
20
11
11 sin 115°
sin B =
20
B = 29.9°
Law of sines
0.4985 Multiply each side by 11.
Use inverse sine function.
You then know that C 180° – 115° – 29.9° = 35.1°. Use
the law of sines again to find the remaining side
length c of the triangle.
c
sin 35.1° =
20
Law of sines
sin 115°
20 sin 35.1°
c =
sin 115° Multiply each side by sin 35.1°.
c
12.7
Use a calculator.
ANSWER
In ABC, B
29.9°, C
35.1°, and c
12.7.
This is a SSA case with one solution.
Solve
ABC with A = 51°, a = 3.5, and b = 5.
SOLUTION
Begin by drawing a horizontal line. On one end form a 51°
angle (A) and draw a segment 5 units long (AC , or b). At
vertex C, draw a segment 3.5 units long (a). You can see that
a needs to be at least 5 sin 51°
3.9 units long to reach the
horizontal side and form a triangle.
So, it is not possible to draw the indicated triangle.
SSA case with no solution
Solve
ABC with A = 40°, a = 13, and b = 16.
SOLUTION
First make a sketch. Because b sin A = 16 sin 40° 10.3,
and 10.3 < 13 < 16 (h < a < b), two triangles can be formed.
Triangle 1
Triangle 2
See Example 4, page 588
Use the law of sines to find the possible measures of B.
sin B
sin 40°
=
16
13
16 sin 40°
sin B =
13
Law of sines
0.7911 Use a calculator.
There are two angles B between 0° and 180° for which
sin B
0.7911. One is acute and the other is obtuse.
Use your calculator to find the acute angle: sin–1 0.7911
52.3°.
The obtuse angle has 52.3° as a reference angle, so
its measure is 180° – 52.3° = 127.7°. Therefore, B 52.3°
or B 127.7°.
Now find the remaining angle C and side length c for
each triangle.
Triangle 1
C
Triangle 2
180° – 40° – 52.3° = 87.7°
C
13
c
=
sin 40°
sin 87.7°
13 sin 87.7°
c =
sin 40°
ANSWER
180° – 40° – 127.7° = 12.3°
13
c
=
sin 40°
sin 12.3°
20.2
c =
13 sin 12.3°
sin 40°
4.3
ANSWER
In Triangle 1, B 52.3°,
C 87.7°, and c 20.2.
In Triangle 2, B 127.7°,
C 12.3°, and c 4.3.
SSA case with two solutions
Solve
ABC.
3.
A = 122°, a = 18, b = 12
SOLUTION
sin B
sin 122°
=
Law of sines
18
12
12 sin 122°
0.5653 Multiply each side by 12.
sin B =
18
Use inverse sine function.
B = 34.4°
You then know that C 180° – 122° – 34.4° = 23.6°. Use
the law of sines again to find the remaining side
length c of the triangle.
c
18
Law of sines
sin 23.6° = sin 122°
18 sin 23.6°
c =
sin 122° Multiply each side by sin 23.6°.
Use a calculator.
c
8.5
ANSWER
In ABC, B
34.4°, C
23.6°, and c
8.5.
Solve
ABC.
4.
A = 36°, a = 9, b = 12
SOLUTION
Because b sin A = 12 sin 36° ≈ 7.1, and 7.1 < 9 < 13 (h < a < b),
two triangles can be formed.
Use the law of sines to find the possible measures of B.
sin B
sin 36°
=
Law of sines
12
9
12 sin 36°
0.7837 Use a calculator.
sin B =
9
There are two angles B between 0° and 180° for which
sin B
0.7831. One is acute and the other is obtuse.
Use your calculator to find the acute angle: sin–1 0.7831
51.6°.
The obtuse angle has 51.6° as a reference angle, so
its measure is 180° – 51.6° = 128.4°. Therefore, B 51.6°
or B 128.4°.
Now find the remaining angle C and side length c for
each triangle.
Triangle 1
C
Triangle 2
180° – 36° – 51.6° = 92.4°
C
9
c
=
sin 36°
sin 92.4°
9 sin 92.4°
c =
sin 36°
ANSWER
In Triangle 1, B 51.6°,
C 82.4°, and c 15.3.
180° – 36° – 128.4° = 15.6°
9
c
=
sin 36°
sin 15.6°
15.3
c =
9 sin 15.6°
sin 36°
4
ANSWER
In Triangle 2, B 128.4°,
C 15.6°, and c 4.
Solve ABC.
5. A = 50°, a = 2.8, b = 4
2.8 ? b · sin A
2.8 ? 4 · sin 50°
2.8 < 3.06
ANSWER
Since a is less than 3.06, based on the law of
sines, these values do not create a triangle.
ABC.
Solve
6.
A = 105°, b = 13, a = 6
SOLUTION
sin A
sin 105°
=
Law of sines
13
6
6 sin 105°
0.4458 Multiply each side by 6.
sin A =
13
Use inverse sine function.
A = 26.5°
You then know that C 180° – 105° – 26.5° = 48.5°. Use
the law of sines again to find the remaining side
length c of the triangle.
c
13
=
Law of sines
sin 48.5°
sin 105°
13 sin 48.5°
c =
sin 105° Multiply each side by sin 48.5°.
c
ANSWER
10.1
Use a calculator.
In ABC, A
26.5°, C 48.5°, and c
10.1.
When can the law of sines be used to solve a
triangle?
• The law of sines says that in any triangle ABC,
sin 𝐴
sin 𝐵
𝑠𝑖𝑛 𝐶
=
=
.
𝑎
𝑏
𝑐
• The law of sines is used to solve triangles with no
right angle in the AAS, ASA and SSA cases.
How is the SSA case different from the AAS and ASA
cases?
• In the SSA case, there may be one triangle, two
triangles, or no triangles with the given
measurements.
9.5 Assignment, day 1
Page 590, 3-25 odd
No work is the same as a missing problem