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?

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).

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

asin 48° sin 107°

c=15sin 25°=

Write two equations, each with one variable.

asin 48°

15sin 25°= sin 107°

c 15sin 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 = 8SOLUTION

By the law of sines, you can write

asin 46° sin 100°

c= 8sin 34°=

Write two equations, eachwith one variable.

asin 46°

8sin 34°= sin 100°

c 8sin 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

c8 sin 100°

sin 34°=a8 sin 46°

sin 34°=

2. A = 51°, B = 44°, c = 11Solve ABC.

SOLUTION

By the law of sines, you can write

asin 51° sin 85°

11= bsin 44°=

Write two equations, eachwith one variable.

asin 51°

11sin 85°= sin 44°

b 11sin 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 b11 sin 44°

sin 85°=11 sin 51°

sin 85°=

SSA CaseTwo 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 B11

= sin 115°20

Law of sines

sin B =11 sin 115°

200.4985 Multiply each side by 11.

B = 29.9° 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.

Law of sinesc

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). 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.

Law of sinessin B16 = sin 40°

13

sin B =16 sin 40°

130.7911 Use a calculator.

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°.

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°.

Now find the remaining angle C and side length c for each triangle.

C 180° – 40° – 52.3° = 87.7° C 180° – 40° – 127.7° = 12.3°

csin 87.7° =

13sin 40°

csin 12.3° =

13sin 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 B12

= sin 122°18

Law of sines

sin B =12 sin 122°

180.5653 Multiply each side by 12.

B = 34.4° Use inverse sine function.

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.

SOLUTION

c sin 23.6°

18sin 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 = 12SOLUTION

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 sinessin B12 = sin 36°

9

sin B =12 sin 36°

90.7837 Use a calculator.

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°.

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°.

Now find the remaining angle C and side length c for each triangle.

C 180° – 36° – 51.6° = 92.4° C 180° – 36° – 128.4° = 15.6°

csin 92.4° =

9sin 36°

csin 15.6° =

9sin 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 = 4Solve ABC.

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.

Solve ABC. 6. A = 105°, b = 13, a = 6

sin A6

= sin 105°13

Law of sines

sin A = 6 sin 105°

130.4458 Multiply each side by 6.

A = 26.5° Use inverse sine function.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.

SOLUTION

c sin 48.5°

13sin 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

When can the law of sines be used to solve a triangle?• The law of sines says that in any triangle ABC, .• 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 oddNo work is the same as a missing problem