Answer
See below
Work Step by Step
We are given $a, b, C$. Use law of cosines to find $c$:
$$a^2=b^2+c^2-2bc\cos A\\ c=\sqrt a^2+b^2-2ab\cos C\\c \approx 19.3$$
Use law of sines to find: $\frac{\sin B}{b}=\frac{\sin C}{c}\\\sin B=\frac{\sin C}{c}\times b\\\arcsin (\sin B)=\arcsin (\frac{\sin C}{c}b)\\B=\arcsin(\frac{\sin C}{c}. b)\\B\approx 80.63^\circ$
Since the sum of the triangle is $180^\circ$, we obtain:
$$A+B+C=180^\circ\\C=180^\circ-A-B\\C=180^\circ -80.63^\circ -65^\circ\\C=34.37^\circ$$
