## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 6 - Section 6.1 - The Law of Sines - Exercise Set - Page 720: 16

#### Answer

$A=90{}^\circ,b\approx 7.9$ and $c\approx 1.4$.

#### 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+80{}^\circ +10{}^\circ =180{}^\circ \\ & A=180{}^\circ -90{}^\circ \\ & A=90{}^\circ \end{align} Now, we will find the remaining sides using the ratio: $\frac{a}{\sin A}$,or $\frac{8}{\sin 90{}^\circ }$, Now, we will use the law of sines to find b: \begin{align} & \frac{b}{\sin B}=\frac{a}{\sin A} \\ & \frac{b}{\sin 80{}^\circ }=\frac{8}{\sin 90{}^\circ } \\ & b=\frac{8\sin 80{}^\circ }{\sin 90{}^\circ } \\ & b\approx 7.9 \end{align} Using the law of sines again, we will find c. \begin{align} & \frac{c}{\sin C}=\frac{a}{\sin A} \\ & \frac{c}{\sin 10{}^\circ }=\frac{8}{\sin 90{}^\circ } \\ & c=\frac{8\sin 10{}^\circ }{\sin 90{}^\circ } \\ & c\approx 1.4 \end{align}

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