## 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: 44

#### Answer

The plane is a distance of 182 miles from $X$

#### Work Step by Step

Suppose that the plane flies from point $X$ to point $Y$ and then to point $Z$. We need to find the distance between $X$ and $Z$. First we can find the angle formed at the point $Y$: The angle from the vertical formed at the point $X$ is $180^{\circ}-125^{\circ} = 55^{\circ}$ The angle $Y = (270^{\circ}-230^{\circ})+(90^{\circ}-55^{\circ}) = 75^{\circ}$ We can use the law of cosines to find the distance $XZ$: $XZ^2 = XY^2+YZ^2-2(XY)(YZ)~cos~Y$ $XZ = \sqrt{XY^2+YZ^2-2(XY)(YZ)~cos~Y}$ $XZ = \sqrt{(180~mi)^2+(100~mi)^2-(2)(180~mi)(100~mi)~cos~75^{\circ}}$ $XZ = \sqrt{33082.5~mi^2}$ $XZ = 182~mi$ The plane is a distance of 182 miles from $X$

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