Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 8 - Application of Trigonometry - 8.2 The Law of Cosines - 8.2 Exercises: 13

Answer

$ \left| CB\right| \approx 7.032$ $ \angle C\approx 21.449$ $\angle B\approx 34.551$

Work Step by Step

From the laws of cosines $\left| CB\right| ^{2}=5^{2}+3^{2}-2\times 5\times 3\times \cos 121\approx 49.451\Rightarrow \left| CB\right| \approx 7.032$ From the law of sines $\dfrac {\left| CB\right| }{\sin 121}=\dfrac {3}{\sin \angle C}\Rightarrow \sin \angle C=\dfrac {3\sin 121}{\left| CB\right| }\approx 0.365\Rightarrow \angle C\approx 21.449$ Internal angles of a triangle is $180^0$ So $\angle A+\angle B+\angle C=180\Rightarrow \angle B=180-\left( \angle A+\angle C\right) \approx 180-\left( 121+21.449\right) \approx 34.551$
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