## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 6 - Section 6.1 - The Law of Sines - Exercise Set - Page 720: 14

#### Answer

$A=50{}^\circ,a\approx 1757.9$ and $c\approx 1879.7$.

#### Work Step by Step

First we will find the value of A. Properties of a triangle: The sum of three angles is $A+B+C=180{}^\circ$ \begin{align} & A+B+C=180{}^\circ \\ & A+5{}^\circ +125{}^\circ =180{}^\circ \\ & A=180{}^\circ -130{}^\circ \\ & A=50{}^\circ \end{align} Now, we will find the remaining sides using the ratio $\frac{b}{\sin B}$,or $\frac{200}{\sin 5{}^\circ }$, Now, we will use the law of sines to find a. \begin{align} & \frac{a}{\sin A}=\frac{b}{\sin B} \\ & \frac{a}{\sin 50{}^\circ }=\frac{200}{\sin 5{}^\circ } \\ & a=\frac{200\sin 50{}^\circ }{\sin 5{}^\circ } \\ & a\approx 1757.9 \end{align} Using the law of sines again, we will find c. \begin{align} & \frac{c}{\sin C}=\frac{b}{\sin B} \\ & \frac{c}{\sin 125{}^\circ }=\frac{200}{\sin 5{}^\circ } \\ & c=\frac{200\sin 125{}^\circ }{\sin 5{}^\circ } \\ & c\approx 1879.7 \end{align}

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