Solve the triangle. Round lengths to the nearest tenth and angle measures to the nearest degree. a = 6, c = 11, B = 109° A. b = 14.1, A = 24°, C = 47° B. b = 19.9, A = 22°, C = 49° C. b = 17, A = 26°, C = 45° D. no triangle

Question

Desmond Moses

Mathematics

20 June, 12:31

Solve the triangle. Round lengths to the nearest tenth and angle measures to the nearest degree. a = 6, c = 11, B = 109° A. b = 14.1, A = 24°, C = 47° B. b = 19.9, A = 22°, C = 49° C. b = 17, A = 26°, C = 45° D. no triangle

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Answers (1)

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Rylee Clay · Answer 1

Use cosine rule to find b

b^2 = 6^2 + 11^2 - 2*6*11cos 109 = 199.97

b = 14.1 to nearest tenth.

now use sine rule to find the value of < A

14.14 / sin 109 = 6 / sin A

Its A