## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 7 - Test - Page 353: 7

#### Answer

The angles of the triangle are as follows: $A = 60^{\circ}, B = 40.9^{\circ},$ and $C = 79.1^{\circ}$ The lengths of the sides are as follows: $a = 39.7~m, b = 30~m,$ and $c = 45~m$

#### Work Step by Step

$A = 60^{\circ}$ $b = 30~m$ $c = 45~m$ We can use the law of cosines to find $a$: $a^2 = b^2+c^2-2bc~cos~A$ $a = \sqrt{b^2+c^2-2bc~cos~A}$ $a = \sqrt{(30~m)^2+(45~m)^2-(2)(30~m)(45~m)~cos~60^{\circ}}$ $a = \sqrt{1575~m^2}$ $a = 39.7~m$ We can use the law of sines to find angle $B$: $\frac{a}{sin~A} = \frac{b}{sin~B}$ $sin~B = \frac{b~sin~A}{a}$ $B = arcsin(\frac{b~sin~A}{a})$ $B = arcsin(\frac{30~sin~60^{\circ}}{39.7})$ $B = arcsin(0.6544)$ $B = 40.9^{\circ}$ We can find angle $C$: $A+B+C = 180^{\circ}$ $C = 180^{\circ}-A-B$ $C = 180^{\circ}-60^{\circ}-40.9^{\circ}$ $C = 79.1^{\circ}$

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