Answer
See below
Work Step by Step
We are given $a, c, B$. Use law of cosines to find $c$:
$$a^2=b^2+c^2-2bc\cos A\\ c=\sqrt xa^2+b^2-2bc\cos A\\c=\sqrt 29^2+38^2-2(29)(38)\cos 63^\circ\\b\approx 35.84$$
Use law of sines to find: $\frac{\sin B}{b}=\frac{\sin A}{a}\\\sin A=\frac{\sin B}{b}\times a\\\arcsin (\sin A)=\arcsin (\frac{\sin B}{b}a)\\A=\arcsin(\frac{\sin B}{b}. a)\\A\approx 46.1^\circ$
Since the sum of the triangle is $180^\circ$, we obtain:
$$A+B+C=180^\circ\\C=180^\circ-A-B\\C=180^\circ -46.1^\circ -63^\circ\\C=70.9^\circ$$
