## Precalculus (6th Edition) Blitzer

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