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

Answer

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 $\begin{align} & {{c}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\ \cos C \\ & {{c}^{2}}={{15}^{2}}+{{10}^{2}}-2\left( 15 \right)\left( 10 \right)\cos 145{}^\circ \\ & {{c}^{2}}=\pm \sqrt{570.7} \\ & c\simeq 23.9 \end{align}$
Small 1569806350
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.