## Precalculus (6th Edition) Blitzer

$B\approx51^\circ$, $C\approx94^\circ$, $c\approx19.1$
Step 1. Based on the given conditions, using the Law of Sines, we have $\frac{sinB}{15}=\frac{sin35^\circ}{11}$ Thus $sinB\approx0.7821$ and $B=asin(0.7821)\approx51^\circ$ (acute) Step 2. We can find the angle as $C=180-35-51\approx94^\circ$ Step 3. Using the Law of Sines, we have $\frac{sinC}{c}=\frac{sin35^\circ}{11}$ Thus $c=\frac{11sin94^\circ}{sin35^\circ}\approx19.1$