9.5 Apply the Law of Sines
When can the law of sines be used tosolve a triangle?
How is the SSA case different from theAAS and ASA cases?
The law of sines can be used to solve triangles when twoangles and the length of any side are known. (AAS or ASAcases), or when the lengths of two sides and an angle oppositeone of the two side are known (SSA case).
Law of Sines
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
sin 48°
sin 107°
c
=
15
sin 25°
=
Write two equations,
each
with one variable.
a
sin 48°
15
sin 25°
=
sin 107°
c
15
sin 25°
=
Solve for each variable.
a =
15 sin 48°
sin 25°
c =
15 sin 107°
sin 25°
Use a calculator.
a
26.4
c
33.9
In ABC, A = 48°, a 26.4, and c 33.9.
ANSWER
Solve ABC.
1. B = 34°, C = 100°, b = 8
SOLUTION
By the law of sines, you can write
a
sin 46°
sin 100°
c
=
8
sin 34°
=
Write two equations, each
with one variable.
a
sin 46°
8
sin 34°
=
sin 100°
c
8
sin 34°
=
First find the angle:
A = 180° – 34° – 100°
= 46°.
Solve for each variable.
Use a calculator.
a
10.3
c
14.1
In ABC, A 46°, a 10.3, and c 14.1.
ANSWER
c
8 sin 100°
sin 34°
=
a
8 sin 46°
sin 34°
=
2. A = 51°, B = 44°, c = 11
Solve ABC.
SOLUTION
By the law of sines, you can write
a
sin 51°
sin 85°
11
=
b
sin 44°
=
Write two equations, each
with one variable.
a
sin 51°
11
sin 85°
=
sin 44°
b
11
sin 85°
=
First find the angle:
C = 180° – 51° – 44°
= 85°.
Solve for each variable.
Use a calculator.
a
8.6
b
7.7
In ABC, A 85°, a 8.6, and b 7.7.
ANSWER
a
b
11 sin 44°
sin 85°
=
11 sin 51°
sin 85°
=
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 sideopposite A is longer than the given adjacent side, youknow that only one triangle can be formed. Use the law ofsines to find B.
sin B
11
=
sin 115°
20
Law of sines
sin B
=
11 sin 115°
20
0.4985
Multiply each side by 11.
B
=
29.9°
Use inverse sine function.
You then know that C 180° – 115° – 29.9° = 35.1°. Usethe law of sines again to find the remaining sidelength c of the triangle.
Law of sines
c
sin 35.1°
20
sin 115°
=
c
=
20 sin 35.1°
sin 115°
Multiply each side by sin 35.1°.
c
12.7
Use a calculator.
In ABC, B 29.9°, C 35.1°, and c 12.7.
ANSWER
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). Atvertex C, draw a segment 3.5 units long (a). You can see thata needs to be at least 5 sin 51° 3.9 units long to reach thehorizontal 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.
Law of sines
sin B
16
=
sin 40°
13
sin B
=
16 sin 40°
13
0.7911
Use a calculator.
The obtuse angle has 52.3° as a reference angle, soits measure is 180° – 52.3° = 127.7°. Therefore, B 52.3°or B 127.7°.
There are two angles B between 0° and 180° for whichsin 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°.
Now find the remaining angle C and side length c foreach triangle.
C 180° – 40° – 52.3° = 87.7°
C 180° – 40° – 127.7° = 12.3°
c
sin 87.7°
=
13
sin 40°
c
sin 12.3°
=
13
sin 40°
c
=
13 sin 87.7°
sin 40°
20.2
c
=
13 sin 12.3°
sin 40°
4.3
Triangle 1
Triangle 2
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.
ANSWER
ANSWER
SSA case with two solutions
Solve ABC.
3. A = 122°, a = 18, b = 12
sin B
12
=
sin 122°
18
Law of sines
sin B
=
12 sin 122°
18
0.5653
Multiply each side by 12.
B
=
34.4°
Use inverse sine function.
You then know that C 180° – 122° – 34.4° = 23.6°. Usethe law of sines again to find the remaining sidelength c of the triangle.
SOLUTION
c
sin 23.6°
18
sin 122°
=
Law of sines
c
=
18 sin 23.6°
sin 122°
Multiply each side by sin 23.6°.
c
8.5
Use a calculator.
In ABC, B 34.4°, C 23.6°, and c 8.5.
ANSWER
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.
Law of sines
sin B
12
=
sin 36°
9
sin B
=
12 sin 36°
9
0.7837
Use a calculator.
The obtuse angle has 51.6° as a reference angle, soits measure is 180° – 51.6° = 128.4°. Therefore, B 51.6°or B 128.4°.
There are two angles B between 0° and 180° for whichsin 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°.
Now find the remaining angle C and side length c foreach triangle.
C 180° – 36° – 51.6° = 92.4°
C 180° – 36° – 128.4° = 15.6°
c
sin 92.4°
=
9
sin 36°
c
sin 15.6°
=
9
sin 36°
c
=
9 sin 92.4°
sin 36°
15.3
c
=
9 sin 15.6°
sin 36°
4
Triangle 1
Triangle 2
In Triangle 1, B 51.6°,
C 82.4°,
and c 15.3.
In Triangle 2, B 128.4°,
C 15.6°,
and c 4.
ANSWER
ANSWER
2.8 ? b · sin A
5. A = 50°, a = 2.8, b = 4
Solve ABC.
2.8 ? 4 · sin 50°
2.8 < 3.06
ANSWER
Since a is less than 3.06, based on the law ofsines, these values do not create a triangle.
Solve ABC.
6. A = 105°, b = 13, a = 6
sin A
6
=
sin 105°
13
Law of sines
sin A
=
6 sin 105°
13
0.4458
Multiply each side by 6.
A
=
26.5°
Use inverse sine function.
You then know that C 180° – 105° – 26.5° = 48.5°. Usethe law of sines again to find the remaining sidelength c of the triangle.
SOLUTION
c
sin 48.5°
13
sin 105°
=
Law of sines
Multiply each side by sin 48.5°.
c
=
13 sin 48.5°
sin 105°
c
10.1
Use a calculator.
In ABC, A 26.5°, C 48.5°, and c 10.1.
ANSWER
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