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: 58

Answer

$ N 82.5^{\circ} W $

Work Step by Step

The trigonometric ratios are as follows: $\sin \theta= \dfrac{opposite}{hypotenuse}$ ; $\cos \theta= \dfrac{Adjacent}{hypotenuse}$ and $\tan \theta= \dfrac{Opposite}{Adjacent}$ Since, $\tan \theta =\dfrac{11}{7}$ This implies that $\theta =\tan^{-1} \dfrac{11}{7}=57.5^{\circ}$ So, the bearing measurement is: $180^{\circ}-57.5^{\circ}-40^{\circ}=82.5^{\circ}$ So, our required answer is: $ N 82.5^{\circ} W $
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