1. 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?

2. 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

3. 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 a sin 48° sin 107° c = 15 sin 25° = Write two equations, each with one variable. a sin 48° 15 sin 25° = sin 107° c 15 sin 25° = Solve for each variable. a = 15 sin 48° sin 25° c = 15 sin 107° sin 25° Use a calculator. a26.4c33.9 In ABC, A = 48°, a 26.4, and c 33.9. ANSWER

4. Solve ABC. 1. B = 34°, C = 100°, b = 8 SOLUTION By the law of sines, you can write a sin 46° sin 100° c = 8 sin 34° = Write two equations, each with one variable. a sin 46° 8 sin 34° = sin 100° c 8 sin 34° = First find the angle: A = 180° – 34° – 100° = 46°. Solve for each variable. Use a calculator. a10.3c14.1 In ABC, A 46°, a 10.3, and c 14.1. ANSWER c 8 sin 100° sin 34° = a 8 sin 46° sin 34° =

5. 2. A = 51°, B = 44°, c = 11 Solve ABC. SOLUTION By the law of sines, you can write a sin 51° sin 85° 11 = b sin 44° = Write two equations, each with one variable. a sin 51° 11 sin 85° = sin 44° b 11 sin 85° = First find the angle: C = 180° – 51° – 44° = 85°. Solve for each variable. Use a calculator. a8.6b7.7 In ABC, A 85°, a 8.6, and b 7.7. ANSWER ab 11 sin 44° sin 85° = 11 sin 51° sin 85° =

6. SSA Case Two 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.

7. 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 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.

8. 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 This is a SSA case with one solution.

9. 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

10. 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 See Example 4, page 588

11. 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°.

12. 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 SSA case with two solutions

13. 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 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

14. 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. 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°.

15. 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

16. 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.

17. 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 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

19. 9.5 Assignment, day 1 Page 590, 3-25 odd No work is the same as a missing problem