Answer
$c\approx3.7, A\approx52^\circ, B\approx33^\circ, C=95^\circ$
Work Step by Step
Step 1. From the given configuration, we have
$\angle ACB=C=180^\circ-35^\circ-50^\circ=95^\circ$
Step 2. Using the Law of Cosines, we have $c^2=a^2+b^2-2ab\ cos(C)=3^2+2^2-2(3)(2)cos(95^\circ)\approx 14.046$, thus $c\approx3.7$
Step 3. Using the Law of Sines, we have
$\frac{sinB}{b}=\frac{sin(C}{c}$, thus $sinB=\frac{2sin(95^\circ)}{3.7}\approx0.5385$ and $B=asin(0.5385)\approx33^\circ$, and $A\approx180^\circ-95^\circ-33^\circ=52^\circ$
Step 4. We have the solutions as
$c\approx3.7, A\approx52^\circ, B\approx33^\circ, C=95^\circ$
