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


$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.
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.