## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 6 - Section 6.1 - The Law of Sines - Exercise Set - Page 720: 10

#### Answer

$B=100{}^\circ,b\approx 26.1$ and $c\approx 10.8$.

#### Work Step by Step

First we will find the value of B Properties of a triangle: Sum of three angles is $A+B+C=180{}^\circ$ \begin{align} & A+B+C=180{}^\circ \\ & 56{}^\circ +B+24{}^\circ =180{}^\circ \\ & B=180{}^\circ -80{}^\circ \\ & B=100{}^\circ \end{align} Now, we will find the remaining sides using the ratio: $\frac{a}{\sin A}$,or $\frac{22}{\sin 56{}^\circ }$, Now, we will use the law of sines to find b. \begin{align} & \frac{b}{\sin B}=\frac{a}{\sin A} \\ & \frac{b}{\sin 100{}^\circ }=\frac{22}{\sin 56{}^\circ } \\ & b=\frac{22\sin 100{}^\circ }{\sin 56{}^\circ } \\ & b\approx 26.1 \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 24{}^\circ }=\frac{22}{\sin 56{}^\circ } \\ & c=\frac{22\sin 24{}^\circ }{\sin 56{}^\circ } \\ & c\approx 10.8 \end{align}

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