## 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 321: 43

#### Answer

C is a distance of 281.3 km from A

#### Work Step by Step

The points A, B, and C form a triangle. The angle $B = 90^{\circ}- 51^{\circ}20'$ which is $38^{\circ}40'$ We can use the law of cosines to find the distance $AC$: $AC^2 = AB^2+BC^2-2(AB)(BC)~cos~B$ $AC = \sqrt{AB^2+BC^2-2(AB)(BC)~cos~B}$ $AC = \sqrt{(450~km)^2+(359~km)^2-(2)(450~km)(359~km)~cos~38^{\circ}40'}$ $AC = \sqrt{79107.6~km^2}$ $AC = 281.3~km$ C is a distance of 281.3 km from A

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