## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 9 - Section 9.5 - Parametric Equations - Exercise Set - Page 1022: 96

#### Answer

The third side of the triangle is $4.4$ units long and the other two angles are $45{}^\circ$ and $96{}^\circ$.

#### Work Step by Step

We use the cosine law to find out the side of the triangle: \begin{align} & {{a}^{2}}={{b}^{2}}+{{c}^{2}}-2bc\cos A \\ & {{a}^{2}}={{5}^{2}}+{{7}^{2}}-2.5.7\cos 39{}^\circ \\ & =19.5998 \end{align} The value of the third side is as given below: \begin{align} & a=\sqrt{19.5998} \\ & \approx 4.4 \end{align} Then, to calculate angles, apply the law of sines \begin{align} & \frac{a}{\sin A}=\frac{b}{\sin B} \\ & \frac{4.4}{\sin 39{}^\circ }=\frac{5}{\sin B} \\ & \sin B=\frac{5\sin 39{}^\circ }{4.4} \\ & B\approx 45{}^\circ \end{align} Now, the third angle would be $180-39-45=96{}^\circ$ Thus, the third side of the tringle is 4.4 units long and the other two angles are $45{}^\circ$ and $96{}^\circ$.

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