## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 6 - Section 6.1 - The Law of Sines - Exercise Set - Page 720: 18

#### Answer

There is one triangle and the solution is ${{B}_{1}}\left( \text{or }B \right)\approx 31{}^\circ,\ C\approx 99{}^\circ,\ \text{ and }\ c\approx 38.7$.

#### Work Step by Step

At first we will use the law of sines to find B: \begin{align} & \frac{a}{\sin A}=\frac{b}{\sin B} \\ & \frac{30}{\sin 50{}^\circ }=\frac{20}{\sin B} \\ & \sin B=\frac{20\sin 50{}^\circ }{30} \\ & \approx 0.5107 \end{align} There are two angles between 0° and 180° for which $\sin B=0.5107$ is possible: ${{B}_{1}}\approx 31{}^\circ$ Since we know that the sine function is positive in the second quadrant, the second angle is \begin{align} & {{B}_{2}}\approx 180{}^\circ -31{}^\circ \\ & \approx 149{}^\circ \end{align} Here, ${{B}_{2}}$ is impossible, as $50{}^\circ +149{}^\circ =199{}^\circ$. Now, we will find $C$ using ${{B}_{1}}$ and $A=50{}^\circ$. Using the angle sum property, we will find C: \begin{align} & C=180{}^\circ -{{B}_{1}}-A \\ & \approx 180{}^\circ -31{}^\circ -50{}^\circ \\ & \approx 99{}^\circ \end{align} Using the law of sines, we get side c: \begin{align} & \frac{c}{\sin C}=\frac{a}{\sin A} \\ & \frac{c}{\sin 99{}^\circ }=\frac{30}{\sin 50{}^\circ } \\ & c=\frac{30\sin 99{}^\circ }{\sin 50{}^\circ } \\ & \approx 38.7 \end{align}

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