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Solve the triangle. Round the side length to the nearest tenth and the angle measures to the nearest degree.

Solve the triangle. Round the side length to the nearest tenth and the angle measures-example-1

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Answer: 41, 54, 6.1

Explanation:

By the law of Cosines,


b^2=4^2 +5^2 -2(4)(5)\cos 85^(\circ)\\\\b=\sqrt{4^2 +5^2 -2(4)(5)\cos 85^(\circ)} \\\\ =\sqrt{41-40 \cos 85^(\circ)}\\\\\approx 6.1

By the law of sines,


(\sin A)/(a)=(\sin C)/(c)=(\sin B)/(b)\\\\(\sin A)/(4)=(\sin C)/(5)=\frac{\sin 85^(\circ)}{\sqrt{41-40 \cos (85^(\circ))}}\\\\\implies \sin A=\frac{4\sin 85^(\circ)}{\sqrt{41-40 \cos 85^(\circ)}} \implies \angle A \approx 41^(\circ)\\\\\implies \sin C=\frac{5\sin 85^(\circ)}{\sqrt{41-40 \cos 85^(\circ)}} \implies \angle C \approx 54^(\circ)

User Govind Kumawat
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