Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Section 4.8 - Applications of Trigonometric Functions - Exercise Set - Page 639: 51

Answer

$90$ miles north, $120$ miles east

Work Step by Step

The trigonometric ratios are as follows: $\sin \theta= \dfrac{opposite}{hypotenuse}$ ; $\cos \theta= \dfrac{Adjacent}{hypotenuse}$ and $\tan \theta= \dfrac{Opposite}{Adjacent}$ We can use the sine and cosine of the triangle to solve for the distance. Since, $\cos \theta= \dfrac{Adjacent}{hypotenuse}$ Now, miles to north $ = 150 \times \cos 53^{\circ}=90$ miles and $\sin \theta= \dfrac{opposite}{hypotenuse}$ Now, miles to east $ = 150 \times \sin 53^{\circ}=120$ miles
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