## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 7 - Test - Page 353: 3

#### Answer

$49^{\circ}$

#### Work Step by Step

We can use the law of cosines here because we know the lengths of three sides of the triangle and we need to find the measure of the angle. The law of cosines is: $b^{2}=a^{2}+c^{2}-2ac\cos B$ where $a,b,c$ are the two known sides of the triangle while $B$ is the unknown angle. Substituting the values in the formula and solving: $b^{2}=a^{2}+c^{2}-2ac\cos B$ $22.6^{2}=17.3^{2}+29.8^{2}-2(17.3)(29.8)\cos B$ $510.76=299.29+888.04-1031.08\cos B$ $510.76=1187.33-1031.08\cos B$ $510.76-1187.33=-1031.08\cos B$ $-676.57=-1031.08\cos B$ $-1031.08\cos B=-676.57$ $\cos B=\frac{-676.57}{-1031.08}$ $\cos B=\frac{676.57}{1031.08}$ $B=\cos^{-1} \frac{676.57}{1031.08}$ $B=48.99^{\circ}\approx49^{\circ}$

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