EXAMPLE 2 Solve the SSA case with one solution

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.

EXAMPLE 2 Solve the SSA case with one solution

sin B11

=sin 115°

20Law 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.

EXAMPLE 2 Solve the SSA case with one solution

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

EXAMPLE 3 Examine the SSA case with no 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.

EXAMPLE 4 Solve the SSA case with two solutions

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

EXAMPLE 4 Solve the SSA case with two solutions

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°.

EXAMPLE 4 Solve the SSA case with two solutions

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°

=13

sin 40°c

sin 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

GUIDED PRACTICE for Examples 2, 3, and 4

Solve ABC.

3. A = 122°, a = 18, b = 12

sin B12

=sin 122°

18Law 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

GUIDED PRACTICE for Examples 2, 3, and 4

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

GUIDED PRACTICE for Examples 2, 3, and 4

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.

EXAMPLE 4 Solve the SSA case with two solutions

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°.

EXAMPLE 4 Solve the SSA case with two solutions

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°

=9

sin 36°c

sin 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

GUIDED PRACTICE for Examples 2, 3, and 4

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 of sines, these values do not create a triangle.

GUIDED PRACTICE for Examples 2, 3, and 4

Solve ABC.

6. A = 105°, b = 13, a = 6

sin A6

=sin 105°

13Law 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

GUIDED PRACTICE for Examples 2, 3, and 4

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