Answer
See below
Work Step by Step
We are given $a, b,c$. Use law of cosines to find $A$:
$$a^2=b^2+c^2-2bc\cos A\\ 38^2=31^2+35^2-2(31)(35)\cos A\\\cos A=\frac{38^2-31^2-35^2}{-2(31)(35)}\\A\approx70^\circ$$
Use law of sines to find: $\frac{\sin B}{b}=\frac{\sin A}{a}\\\sin B=\frac{\sin A}{a}\times b\\\arcsin (\sin B)=\arcsin (\frac{\sin A}{a}b)\\B=\arcsin(\frac{\sin A}{a}. b)\\B\approx 50^\circ$
Since the sum of the triangle is $180^\circ$, we obtain:
$$A+B+C=180^\circ\\C=180^\circ-A-B\\C=180^\circ -70^\circ - 50^\circ\\C\approx 60^\circ$$
