Precalculus (6th Edition) Blitzer

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

Chapter 5 - Review Exercises - Page 709: 69

Answer

$12^\circ $ and $78^\circ $

Work Step by Step

Step 1. Using the formula $d=\frac{v_0^2}{16}sin\theta cos\theta$ with $d=100, v_0=90$, we have $100=\frac{90^2}{16}sin\theta cos\theta$ and $2sin\theta cos\theta\approx0.3951$ or $sin2\theta=0.3951$ Step 2. The solutions in $[0,2\pi)$ are $2\theta=sin^{-1}0.3951\approx0.4061\ rad\approx23.27^\circ $ and $2\theta=180^\circ -23.27^\circ =156.73^\circ $ Step 3. We need to limit $0\lt\theta\lt90^\circ $, and higher $2\theta$ values than above will give $\theta\gt90^\circ $. Thus we have $\theta=11.64^\circ \approx12^\circ $ and $\theta=78.37^\circ \approx78^\circ $ as our solutions.
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