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 B

11

sin B

=

sin

115° 20 = 11

sin

20 115°

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.

EXAMPLE 2

Solve the SSA case with one solution

c sin

35.1° =

c

= 20

sin

115°

Law of sines

20

sin

35.1°

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

.

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

.

sin B

16 =

sin

40° 13

Law of sines

sin B

= 16

sin

40° 13 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°

.

EXAMPLE 4

Solve the SSA case with two solutions Now find the remaining angle

C

and side length

c

for each triangle.

Triangle 1 Triangle 2

C

180° – 40° – 52.3° = 87.7°

C

180° – 40° – 127.7° = 12.3°

c sin

87.7° = 13

sin

40°

c

= 13

sin

87.7°

sin

40°

ANSWER In Triangle

1

,

B

52.3°

,

C

87.7°

,and

c

20.2.

20.2

c c sin

12.3° = 13

sin

40° = 13

sin

12.3°

sin

40° 4.3

ANSWER In Triangle

2

,

B

127.7°

,

C

12.3°

,and

c

4.3.

GUIDED PRACTICE

for Examples 2, 3, and 4 Solve

ABC

.

3. A = 122°

,

a

= 18

,

b

= 12

SOLUTION

sin B

12 =

sin

122° 18

sin B

= 12

sin

122° 18 0.5653

Law of sines 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.

GUIDED PRACTICE

for Examples 2, 3, and 4

c sin

23.6° =

c

= 18

sin

122°

Law of sines

18

sin

23.6°

sin

122°

Multiply each side by

sin

23.6°

.

c

8.5

Use a calculator.

ANSWER In

ABC

,

B

34.4°

,

C

23.6°

, and

c

8.5

.

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

.

sin B

12 =

sin

36° 9

Law of sines

sin B

= 12

sin

36° 9 0.7837

Use a calculator.

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°

.

EXAMPLE 4

Solve the SSA case with two solutions Now find the remaining angle

C

and side length

c

for each triangle.

Triangle 1 Triangle 2

C

180° – 36° – 51.6° = 92.4°

C

180° – 36° – 128.4° = 15.6°

c sin

92.4° = 9

sin

36°

c

= 9

sin

92.4°

sin

36°

ANSWER In Triangle

1

,

B

51.6°

,

C

82.4°

,and

c

15.3.

15.3

c c sin

15.6° = 9

sin

36° = 9

sin

15.6°

sin

36° 4

ANSWER In Triangle

2

,

B

128.4°

,

C

15.6°

,and

c

4.

GUIDED PRACTICE

for Examples 2, 3, and 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.

GUIDED PRACTICE

for Examples 2, 3, and 4 Solve

ABC

.

6. A = 105°

,

b

= 13

,

a

= 6

SOLUTION

sin A

6 =

sin

105° 13

sin A

= 6

sin

105° 13 0.4458

Law of sines 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.

GUIDED PRACTICE

for Examples 2, 3, and 4

c sin

48.5° =

c

= 13

sin

105°

Law of sines

13

sin

48.5°

sin

105°

Multiply each side by

sin

48.5°

.

c

10.1

Use a calculator.

ANSWER In

ABC

,

A

26.5°

,

C

48.5°

, and

c

10.1

.