Answer
See below
Work Step by Step
We are given $A, b,c$. Use law of cosines to find $A$:
$$a^2=b^2+c^2-2bc\cos A\\ a=\sqrt 15^2+24^2 -2(15)(24)\cos 103^\circ\\a\approx 31.03$$
Use law of sines to find: $\frac{\sin B}{b}=\frac{\sin A}{a}\\\sin B=\frac{\sin A}{a}\times b\\\arcsin (\sin B)=\arcsin (\frac{\sin A}{a}b)\\B=\arcsin(\frac{\sin A}{a}. b)\\B=\arcsin(\frac{\sin 103^\circ}{31.03}. 15)\approx 28.1^\circ$
Since the sum of the triangle is $180^\circ$, we obtain:
$$A+B+C=180^\circ\\C=180^\circ-A-B\\C=180^\circ -103^\circ - 28.1^\circ\\C\approx 48.9^\circ$$
