Answer
See below
Work Step by Step
We are given $a, b, C$. Use law of cosines to find $c$:
$$a^2=b^2+c^2-2bc\cos A\\ c=\sqrt xa^2+b^2-2bc\cos A\\c=\sqrt 17^2+20^2-2(17)(20)\cos 48^\circ\\c\approx15.3$$
Use law of sines to find: $\frac{\sin B}{b}=\frac{\sin A}{a}\\\sin B=\frac{\sin A}{a}\times b\\\arcsin (\sin B)=\arcsin (\frac{\sin A}{a}b)\\B=\arcsin(\frac{\sin A}{a}. b)\\B\approx 76.3^\circ$
Since the sum of the triangle is $180^\circ$, we obtain:
$$A+B+C=180^\circ\\C=180^\circ-A-B\\C=180^\circ -76.3^\circ - 48^\circ\\C\approx 55.7^\circ$$
