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: 57


$ N 89.5^{\circ} E $

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, $\tan \theta =\dfrac{7}{5}$ This implies that $\theta =\tan^{-1}\dfrac{7}{5}=54.5^{\circ}$ So, the bearing measurement is: $180^{\circ}-54.5^{\circ}-36^{\circ}=89.5^{\circ}$ So, our required answer is: $ N 89.5^{\circ} E $
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.