## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 6 - Section 6.1 - The Law of Sines - Exercise Set - Page 720: 39

#### Answer

The value of $h=481.6$.

#### Work Step by Step

In the provided graph \begin{align} & \angle ABC=180{}^\circ -67{}^\circ \\ & =113{}^\circ \\ & \angle ACB=180{}^\circ -43{}^\circ -113{}^\circ \\ & =24{}^\circ \end{align} Using the law of sines we will find $\overline{BC}$ \begin{align} & \frac{\overline{BC}}{\sin 43{}^\circ }=\frac{312}{\sin 24{}^\circ } \\ & \overline{BC}=\frac{312\sin 43{}^\circ }{\sin 24{}^\circ } \\ & \overline{BC}\approx 523.1 \end{align} Now, we will use the law of sines to find h. \begin{align} & \frac{h}{\sin 67{}^\circ }=\frac{523.1}{\sin 90{}^\circ } \\ & h=\frac{523.1\sin 67{}^\circ }{\sin 90{}^\circ } \\ & h\approx 481.6 \end{align} Hence, the value of $h=481.6$.

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