1. 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.

2. EXAMPLE 2 Solve the SSA case with one solution 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°. Use the law of sines again to find the remaining side length c of the triangle.

3. EXAMPLE 2 Solve the SSA case with one solution Law of sines c sin 35.1° 20 sin 115° = c = 20 sin 35.1° sin 115° Multiply each side by sin 35.1°. c12.7 Use a calculator. In ABC, B 29.9°, C 35.1°, and c 12.7. ANSWER

4. 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.

5. 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 1Triangle 2

6. EXAMPLE 4 Solve the SSA case with two solutions 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, 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°.

7. 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° c sin 87.7° = 13 sin 40° c sin 12.3° = 13 sin 40° c = 13 sin 87.7° sin 40° 20.2c = 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

8. GUIDED PRACTICE for Examples 2, 3, and 4 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°. Use the law of sines again to find the remaining side length c of the triangle. SOLUTION

9. GUIDED PRACTICE for Examples 2, 3, and 4 c sin 23.6° 18 sin 122° = Law of sines c = 18 sin 23.6° sin 122° Multiply each side by sin 23.6°. c8.5 Use a calculator. In ABC, B 34.4°, C 23.6°, and c 8.5. ANSWER

10. 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.

11. EXAMPLE 4 Solve the SSA case with two solutions 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, 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°.

12. 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° c sin 92.4° = 9 sin 36° c sin 15.6° = 9 sin 36° c = 9 sin 92.4° sin 36° 15.3c = 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

13. 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.

14. GUIDED PRACTICE for Examples 2, 3, and 4 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°. Use the law of sines again to find the remaining side length c of the triangle. SOLUTION

15. GUIDED PRACTICE for Examples 2, 3, and 4 c sin 48.5° 13 sin 105° = Law of sines Multiply each side by sin 48.5°. c = 13 sin 48.5° sin 105° c10.1 Use a calculator. In ABC, A 26.5°, C 48.5°, and c 10.1. ANSWER