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Solve ABC with a = 11 , c = 14 , and B = 34°. b 2 61.7. b 2 61.7. 7.85. EXAMPLE 1. Solve a triangle for the SAS case. SOLUTION. Use the law of cosines to find side length b. b 2 = a 2 + c 2 – 2 ac cos B. Law of cosines.

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**Solve ABCwith a = 11,c = 14, and B = 34°.**b2 61.7 b2 61.7 7.85 EXAMPLE 1 Solve a triangle for the SAS case SOLUTION Use the law of cosines to find side length b. b2 = a2 + c2 – 2ac cosB Law of cosines b2 = 112 + 142 – 2(11)(14) cos 34° Substitute for a, c, and B. Simplify. Take positive square root.**=**sin 34° sinA = 7.85 11 sin B sin A 0.7836 sin A = a b A sin–1 0.7836 51.6° 11 sin 34° The third angle Cof the triangle isC180° – 34° – 51.6° = 94.4°. 7.85 InABC,b 7.85,A 51.68,andC 94.48. ANSWER EXAMPLE 1 Solve a triangle for the SAS case Use the law of sines to find the measure of angle A. Law of sines Substitute for a, b, and B. Multiply each side by 11 and Simplify. Use inverse sine.**SolveABC witha = 12,b = 27, and c = 20.**First find the angle opposite the longest side, AC. Use the law of cosines to solve for B. 272 = 122 + 202 cosB = – 2(12)(20) – 0.3854cosB B cos –1 (– 0.3854) 112.7° EXAMPLE 2 Solve a triangle for the SSS case SOLUTION b2 = a2 + c2 – 2ac cosB Law of cosines 272 = 122 + 202 – 2(12)(20) cosB Substitute. Solve for cosB. Simplify. Use inverse cosine.**sin B**sin A = b a sin A = 12 0.4100 sin A = sin 112.7° 12 sin 112.7° 27 The third angle Cof the triangle isC180° – 24.2° – 112.7° = 43.1°. 27 A sin–1 0.4100 24.2° InABC,A 24.2,B 112.7,andC 43.1. ANSWER EXAMPLE 2 Solve a triangle for the SSS case Now use the law of sines to find A. Law of sines Substitute for a, b, and B. Multiply each side by 12 and simplify. Use inverse sine.**Scientists can use a set of footprints to calculate an**organism’s step angle, which is a measure of walking efficiency. The closer the step angle is to180°, the more efficiently the organism walked. EXAMPLE 3 Use the law of cosines in real life Science The diagram at the right shows a set of footprints for a dinosaur. Find the step angle B.**3162 = 1552 + 1972**cosB = – 2(155)(197) – 0.6062 cos B B cos –1 (– 0.6062) 127.3° ANSWER The step angle Bis about 127.3°. EXAMPLE 3 Use the law of cosines in real life SOLUTION b2 = a2 + c2 – 2ac cosB Law of cosines 3162 = 1552 + 1972 – 2(155)(197) cosB Substitute. Solve for cos B. Simplify. Use inverse cosine.**Find the area of ABC.**b2 57 b2 57 7.55 for Examples 1, 2, and 3 GUIDED PRACTICE 1. a = 8,c = 10,B = 48° SOLUTION Use the law of cosines to find side length b. b2 = a2 + c2 – 2ac cosB Law of cosines b2 = 82 + 102 – 2(8)(10) cos 48° Substitute for a,c, and B. Simplify. Take positive square root.**=**sin 48° sinA = 7.55 8 sin B sin A 0.7874 sin A = a b 8 sin 48° The third angle Cof the triangle isC180° – 48° – 52.2° = 79.8°. 7.55 A sin –1 0.7836 51.6° ANSWER InABC,b 7.55,A 52.2°,andC 94.8°. for Examples 1, 2, and 3 GUIDED PRACTICE Use the law of sines to find the measure of angle A. Law of sines Substitute for a, b, and B. Multiply each side by 8 and simplify. Use inverse sine.**Find the area of ABC.**First find the angle opposite the longest side, AC. Use the law of cosines to solve for B. a = 14,b = 16,c = 9 2. 162 = 142 + 92 cosB = – 2(14)(9) for Examples 1, 2, and 3 GUIDED PRACTICE SOLUTION b2 = a2 + c2 – 2ac cosB Law of cosines 162 = 142 + 92 – 2(14)(9) cosB Substitute. Solve for cos B.**sin B**sin A Bcos–1(– 0.0834) 85.7° b a sin A 14 = sin 85.2° = 16 14sin 85.2° 0.8719 sin A = 16 – 0.0834 cosB for Examples 1, 2, and 3 GUIDED PRACTICE Simplify. Use inverse cosine. Use the law of sines to find the measure of angle A. Law of sines Substitute for a,b, and B. Multiply each side by 14 and simplify.**The third angle Cof the triangle isC180° – 85.2° –**60.7° = 34.1°. InABC,A 60.7°,B 85.2°,andC 34.1°. ANSWER A sin–1 0.8719 60.7° for Examples 1, 2, and 3 GUIDED PRACTICE Use inverse sine.**3352 = 1932 + 1862**cosB = – 2(193)(186) 3. What If? In Example 3, suppose that a = 193 cm, b = 335cm, and c = 186cm. Find the step angle θ. – 0.5592 cos B B cos–1 (– 0.5592) 127° ANSWER The step angle Bis about 124°. for Examples 1, 2, and 3 GUIDED PRACTICE SOLUTION b2 = a2 + c2 – 2ac cosB Law of cosines 3352 = 1932 + 1862 – 2(193)(186) cosB Substitute. Solve for cos B. Simplify. Use inverse cosine.

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