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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