## 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: 22

#### Answer

The angles of the triangle are as follows: $A = 67.3^{\circ}, B = 53.2^{\circ},$ and $C = 59.5^{\circ}$ The lengths of the sides are as follows: $a = 43.7~km, b = 37.9~km,$ and $c = 40.8~km$

#### Work Step by Step

We can use the law of cosines to find $a$: $a^2 = b^2+c^2-2bc~cos~A$ $a = \sqrt{b^2+c^2-2bc~cos~A}$ $a = \sqrt{(37.9~km)^2+(40.8~km)^2-(2)(37.9~km)(40.8~km)~cos~67.3^{\circ}}$ $a = \sqrt{1907.58}$ $a = 43.7~km$ We can use the law of cosines to find $B$: $b^2 = a^2+c^2-2ac~cos~B$ $2ac~cos~B = a^2+c^2-b^2$ $cos~B = \frac{a^2+c^2-b^2}{2ac}$ $B = arccos(\frac{a^2+c^2-b^2}{2ac})$ $B = arccos(\frac{(43.7)^2+(40.8)^2-(37.9)^2}{(2)(43.7)(40.8)})$ $B = arccos(0.5995)$ $B = 53.2^{\circ}$ We can find angle $C$: $A+B+C = 180^{\circ}$ $C = 180^{\circ}-A-B$ $C = 180^{\circ}-67.3^{\circ}-53.2^{\circ}$ $C = 59.5^{\circ}$

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