Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 6 - Section 6.2 - The Law of Cosines - Exercise Set - Page 732: 65


The lengths of the parallelogram’s sides are $8.9\text{ inches}$ and $23.9\text{ inches}$.

Work Step by Step

Let $a$ be the side opposite to the angle $35{}^\circ $ and the other side be $c$. The angle formed by a straight line is $180{}^\circ $. So, we will find out $\angle AOB$ as follows: $\begin{align} & \angle AOB=180{}^\circ -\angle BOC \\ & =180{}^\circ -35{}^\circ \\ & =145{}^\circ \end{align}$ According to the cosine law, ${{a}^{2}}={{b}^{2}}+{{c}^{2}}-2bc\cos \theta $ Side $a$ of the triangle is $\begin{align} & {{a}^{2}}={{15}^{2}}+{{10}^{2}}-2\left( 15 \right)\left( 10 \right)\text{ }\cos 35{}^\circ \\ & {{a}^{2}}=79.3 \\ & a=\pm \sqrt{79.3} \\ & a=8.9 \end{align}$ Side c of the triangle is $\begin{align} & {{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\ \cos C \\ & {{c}^{2}}={{15}^{2}}+{{10}^{2}}-2\left( 15 \right)\left( 10 \right)\cos 145{}^\circ \\ & {{c}^{2}}=\pm \sqrt{570.7} \\ & c\simeq 23.9 \end{align}$
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