Precalculus (6th Edition) Blitzer

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

Chapter 5 - Section 5.5 - Trigonometric Equations - Exercise Set - Page 705: 135


$21^\circ, 69^\circ$

Work Step by Step

Step 1. Given the model function $d=\frac{v^2_0}{16}\ sin\theta\ cos\theta$, with $v_0=90\ ft/sec$ and $d=170\ ft$, we have $\frac{90^2}{16}\ sin\theta\ cos\theta=170$ or $sin(2\theta)=0.6716$ Step 2. With $sin^{-1}(0.6716)\approx42^\circ$, the general solutions for the above equation are $2\theta=360^\circ k+42^\circ$ and $2\theta=360^\circ k+180^\circ-42^\circ$, which gives $\theta=180^\circ k+21^\circ$ and $\theta=180^\circ k+69^\circ$ where $k$ is an integer. Step 3. Within $(0^\circ,90^\circ)$, we have $\theta=21^\circ, 69^\circ$
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