Answer
See below
Work Step by Step
We are given $A, b,c$. Use law of cosines to find $A$:
$$a^2=b^2+c^2-2bc\cos A\\ \cos A=\frac{b^2+c^2-a^2}{2bc}\\A=\arccos \frac{b^2+c^2-a^2}{2bc}\\A=\arccos \frac{23^2+14^2-25^2}{2(23)(14)}\approx81.07^\circ$$
Use law of sines to find: $\frac{\sin B}{b}=\frac{\sin A}{a}\\\sin B=\frac{\sin A}{a}\times b\\\arcsin (\sin B)=\arcsin (\frac{\sin A}{a}b)\\B=\arcsin(\frac{\sin A}{a}. b)\\B=\arcsin(\frac{\sin 81.07^\circ}{25}. 23)\approx 65.35^\circ$
Since the sum of the triangle is $180^\circ$, we obtain:
$$A+B+C=180^\circ\\C=180^\circ-A-B\\C=180^\circ -65.35^\circ - 81.07^\circ\\C\approx 33.58 ^\circ$$
