Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 2 - Acute Angles and Right Triangles - Section 2.5 Further Applications of Right Triangles - 2.5 Exercises - Page 85: 42

Answer

$x = 84.7~m$

Work Step by Step

We can convert the angle $30^{\circ}50'$ to degrees: $\theta = 30^{\circ}50' = (30+\frac{50}{60})^{\circ} = 30.8^{\circ}$ We can find the length $d$ of the line in the middle: $\frac{198.4~m}{d} = cos~\theta$ $d = \frac{198.4~m}{cos(30.8^{\circ})}$ $d = 231.0~m$ The angle opposite $x$ is $21^{\circ}30'$. We can convert this angle to degrees: $21^{\circ}30' = (21+\frac{30}{60})^{\circ} = 21.5^{\circ}$ We can find the length of $x$: $\frac{x}{d} = sin(21.5^{\circ})$ $x = d~sin(21.5^{\circ})$ $x = (231.0~m)~sin(21.5^{\circ})$ $x = 84.7~m$
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.