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9.5 Apply the Law of Sines

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?. Law of Sines.

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9.5 Apply the Law of Sines

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  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. Law of Sines 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).

  3. 15 sin 107° Solve ABCwith C = 107°,B = 25°, and b = 15. a = c = sin 25° c a 26.4 33.9 InABC,A = 48°,a 26.4,andc 33.9. ANSWER 15 sin 48° = = sin 25° a a 15 15 15 c c sin 48° sin 48° sin 25° sin 25° sin 25° sin 107° sin 107° Write two equations, each = = with one variable. SOLUTION First find the angle:A = 180° – 107° – 25° = 48°. By the law of sines, you can write Solve for each variable. Use a calculator.

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

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

  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 ABCwithA = 115°,a = 20, andb = 11. = 11 sin 115° 0.4985 sin B = 20 29.9° B = You then know that C 180° – 115° – 29.9° = 35.1°. Use the law of sines again to find the remaining side length cof the triangle. sinB sin 115° 20 11 SOLUTION First make a sketch. Because Ais 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. Law of sines Multiply each side by 11. Use inverse sine function.

  8. = c = c c sin 35.1° 12.7 ANSWER 20 InABC,B 29.9°,C 35.1°,andc 12.7. sin 115° 20 sin 35.1° sin 115° Law of sines Multiply each side by sin 35.1°. Use a calculator. This is a SSA case with onesolution.

  9. Solve ABCwithA = 51°,a = 3.5, and b = 5. 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 aneeds to be at least 5sin51°3.9 units long to reach the horizontal side and form a triangle. So, it is not possible to draw the indicated triangle. SOLUTION SSA case with no solution

  10. Solve ABCwithA = 40°,a = 13, and b = 16. First make a sketch. Because bsinA = 16sin 40°10.3, and 10.3 < 13 < 16(h < a < b), two triangles can be formed. Triangle 1 Triangle 2 SOLUTION See Example 4, page 588

  11. sinB = 16 16 sin 40° sin 40° 0.7911 = sinB 13 13 There are two angles Bbetween 0° and 180° for which sinB 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°. Use the law of sines to find the possible measures of B. Law of sines Use a calculator.

  12. Triangle 1 Triangle 2 C 180° – 40° – 52.3° = 87.7° C 180° – 40° – 127.7° = 12.3° c c = = sin 87.7° sin 12.3° 13 sin 87.7° 13 sin 12.3° c 20.2 c 4.3 = = sin 40° sin 40° 13 13 sin 40° sin 40° In Triangle 1,B 52.3°, C 87.7°, In Triangle 2,B 127.7°, C 12.3°, ANSWER ANSWER andc 20.2. andc 4.3. Now find the remaining angle Cand side length c for each triangle. SSA case with two solutions

  13. Solve ABC. = c = c = c sin 23.6° 8.5 12 sin 122° 0.5653 = You then know that C 180° – 122° – 34.4° = 23.6°. Use the law of sines again to find the remaining side length cof the triangle. 18 sin B 18 InABC,B 34.4°,C 23.6°,andc 8.5. sin 122° 34.4° = ANSWER sinB sin 122° 18 sin 23.6° B 18 sin 122° 12 3. A = 122°,a = 18,b = 12 SOLUTION Law of sines Multiply each side by 12. Use inverse sine function. Law of sines Multiply each side by sin 23.6°. Use a calculator.

  14. Solve ABC. sinB = 12 12 sin 36° sin 36° 0.7837 = sinB 9 9 There are two angles Bbetween 0° and 180° for which sinB 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°. 4. A = 36°,a = 9,b = 12 SOLUTION Because bsinA = 12sin 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 Use a calculator.

  15. Triangle 1 Triangle 2 C 180° – 36° – 51.6° = 92.4° C 180° – 36° – 128.4° = 15.6° c c = = sin 92.4° sin 15.6° 9 sin 92.4° 9 sin 15.6° c 15.3 c 4 = = sin 36° sin 36° 9 9 sin 36° sin 36° In Triangle 1,B 51.6°, C 82.4°, In Triangle 2,B 128.4°, C 15.6°, ANSWER ANSWER andc 15.3. andc 4. Now find the remaining angle Cand side length c for each triangle.

  16. Solve ABC. ANSWER Since a is less than 3.06, based on the law of sines, these values do not create a triangle. 5. A = 50°,a = 2.8,b = 4 2.8 ? b · sin A 2.8 ? 4 · sin 50° 2.8 < 3.06

  17. = c = c = c sin 48.5° 10.1 6 sin 105° 0.4458 = You then know that C 180° – 105° – 26.5° = 48.5°. Use the law of sines again to find the remaining side length cof the triangle. 13 sin A 13 InABC,A 26.5°,C 48.5°,andc 10.1. sin 105° 26.5° = ANSWER sinA sin 105° 13 sin 48.5° A 13 sin 105° 6 Solve ABC. 6. A = 105°,b = 13,a = 6 SOLUTION Law of sines Multiply each side by 6. Use inverse sine function. Law of sines Multiply each side by sin 48.5°. Use a calculator.

  18. When can the law of sines be used to solve a triangle? • The law of sines says that in any triangle ABC, . • The law of sines is used to solve triangles with no right angle in the AAS, ASA and SSA cases. • How is the SSA case different from the AAS and ASA cases? • In the SSA case, there may be one triangle, two triangles, or no triangles with the given measurements.

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

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