## Precalculus (6th Edition) Blitzer

$C= 44^\circ$ $b \approx14.4$ $c \approx10.5$
Step 1. Given angles $A, B$, we can find angle $C$ as: $C=180^\circ-64^\circ-72^\circ=44^\circ$ Step 2. Using the Law of Sines, we have $\frac{sinB}{b}=\frac{sinA}{a}$, thus $b=\frac{a\ sinB}{sinA}=\frac{13.6\ sin72^\circ}{sin64^\circ}\approx14.4$ Step 3. Using the Law of Sines, we have $\frac{sinC}{c}=\frac{sinA}{a}$, thus $c=\frac{a\ sinC}{sinA}=\frac{13.6\ sin44^\circ}{sin64^\circ}\approx10.5$