## Precalculus (6th Edition) Blitzer

the flagpole is leaning, $88^\circ$
Step 1. See figure. Using the Law of Sines, we have $\frac{sinC}{15}=\frac{sin48^\circ}{16}$; thus $sinC=\frac{15sin48^\circ}{16}\approx 0.6967$ and $C=sin^{-1}0.6967\approx44^\circ$ or $C=180^\circ-44^\circ=136^\circ$ (discard this answer) Step 2. As $44^\circ+48^\circ=92^\circ\gt 90^\circ$, the flagpole is leaning. Step 3. We can find the angle as $B=180^\circ-92^\circ=88^\circ$