Answer
See below
Work Step by Step
We are given $a, c, b$. Use law of cosines to find $c$:
$$a^2=b^2+c^2-2bc\cos A\\ A=\arccos\frac{b^2+c^2-a^2}{2bc}\\A=\arccos\frac{3^2+12^2-10^2}{2(3)(12)} \approx 42.6$$
Use law of sines to find: $\frac{\sin B}{b}=\frac{\sin A}{a}\\\sin B=\frac{\sin A}{a}\times b\\\arcsin (\sin B)=\arcsin (\frac{\sin A}{a}b)\\B=\arcsin(\frac{\sin A}{a}. b)\\A\approx 11.7^\circ$
Since the sum of the triangle is $180^\circ$, we obtain:
$$A+B+C=180^\circ\\C=180^\circ-A-B\\C=180^\circ -11.7^\circ -42.6^\circ\\C=125.7^\circ$$
