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EXAMPLE 2

Solve ABC with A = 115° , a = 20 , and b = 11. EXAMPLE 2. Solve the SSA case with one solution. SOLUTION.

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EXAMPLE 2

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  1. Solve ABCwithA = 115°,a = 20, andb = 11. EXAMPLE 2 Solve the SSA case with one solution 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.

  2. = 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 EXAMPLE 2 Solve the SSA case with one solution Law of sines Multiply each side by 11. Use inverse sine function.

  3. = 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° EXAMPLE 2 Solve the SSA case with one solution Law of sines Multiply each side by sin 35.1°. Use a calculator.

  4. 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. EXAMPLE 3 Examine the SSA case with no solution SOLUTION

  5. 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 EXAMPLE 4 Solve the SSA case with two solutions SOLUTION

  6. 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°. EXAMPLE 4 Solve the SSA case with two solutions Use the law of sines to find the possible measures of B. Law of sines Use a calculator.

  7. 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 12.3° 13 sin 87.7° 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. EXAMPLE 4 Solve the SSA case with two solutions Now find the remaining angle Cand side length c for each triangle.

  8. Solve ABC. = 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 34.4° = sinB sin 122° B 18 12 for Examples 2, 3, and 4 GUIDED PRACTICE 3. A = 122°,a = 18,b = 12 SOLUTION Law of sines Multiply each side by 12. Use inverse sine function.

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

  10. Solve ABC. for Examples 2, 3, and 4 GUIDED PRACTICE 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.

  11. 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°. EXAMPLE 4 Solve the SSA case with two solutions 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° – 36° – 51.6° = 92.4° C 180° – 36° – 128.4° = 15.6° c c = = sin 92.4° sin 15.6° 9 sin 15.6° 9 sin 92.4° 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. EXAMPLE 4 Solve the SSA case with two solutions Now find the remaining angle Cand side length c for each triangle.

  13. Solve ABC. ANSWER Since a is less than 3.06, based on the law of sines, these values do not create a triangle. for Examples 2, 3, and 4 GUIDED PRACTICE 5. A = 50°,a = 2.8,b = 4 2.8 ? b · sin A 2.8 ? 4 · sin 50° 2.8 < 3.06

  14. = 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 26.5° = sinA sin 105° A 13 6 for Examples 2, 3, and 4 GUIDED PRACTICE Solve ABC. 6. A = 105°,b = 13,a = 6 SOLUTION Law of sines Multiply each side by 6. Use inverse sine function.

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

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