## Precalculus (6th Edition) Blitzer

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