## Precalculus (6th Edition) Blitzer

$c\approx19.1, A\approx31^\circ, B\approx19^\circ, C=130^\circ$
Step 1. From the given configuration, we have $\angle ACB=C=180^\circ-15^\circ-35^\circ=130^\circ$ Step 2. Using the Law of Cosines, we have $c^2=a^2+b^2-2ab\ cos(C)=13^2+8^2-2(13)(8)cos(130^\circ)\approx366.70$, thus $c\approx19.1$ Step 3. Using the Law of Sines, we have $\frac{sinB}{b}=\frac{sin(130^\circ)}{19.1}$, thus $sinB=\frac{8sin(130^\circ)}{19.1}\approx0.3209$ and $B=asin(0.3209)\approx19^\circ$, and $A=180^\circ-130^\circ-19^\circ=31^\circ$ Step 4. We have the solutions as $c\approx19.1, A\approx31^\circ, B\approx19^\circ, C=130^\circ$