## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 7 - Applications of Trigonometry and Vectors - Section 7.3 The Law of Cosines - 7.3 Exercises - Page 320: 34

#### Answer

The angles of the triangle are as follows: $A = 4.8^{\circ}, B = 168.2^{\circ},$ and $C = 7.0^{\circ}$ The lengths of the sides are as follows: $a = 15.1~cm, b = 34.1~cm,$ and $c = 19.2~cm$

#### Work Step by Step

We can use the law of cosines to find $b$: $b^2 = a^2+c^2-2ac~cos~B$ $b = \sqrt{a^2+c^2-2ac~cos~B}$ $b = \sqrt{(15.1~cm)^2+(19.2~cm)^2-(2)(15.1~cm)(19.2~cm)~cos~168.2^{\circ}}$ $b = \sqrt{1164.236~cm^2}$ $b = 34.1~cm$ We can use the law of cosines to find $C$: $c^2 = a^2+b^2-2ab~cos~C$ $2ab~cos~C = a^2+b^2-c^2$ $cos~C = \frac{a^2+b^2-c^2}{2ab}$ $C = arccos(\frac{a^2+b^2-c^2}{2ab})$ $C = arccos(\frac{15.1^2+34.1^2-19.2^2}{(2)(15.1)(34.1)})$ $C = arccos(0.99258)$ $C = 7.0^{\circ}$ We can find angle $A$: $A+B+C = 180^{\circ}$ $A = 180^{\circ}-B-C$ $A = 180^{\circ}-168.2^{\circ}-7.0^{\circ}$ $A = 4.8^{\circ}$

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