Answer
See below
Work Step by Step
We are given $a, c, b$. Use law of cosines to find $c$:
$$a^2=b^2+c^2-2bc\cos A\\ B=\arccos\frac{b^2+c^2-a^2}{2bc}\\b=\arccos\frac{20^2+23^2-24^2}{2(23)(20)} \approx 67.4$$
Use law of sines to find: $\frac{\sin B}{b}=\frac{\sin A}{a}\\\sin A=\frac{\sin B}{b}\times a\\\arcsin (\sin A)=\arcsin (\frac{\sin B}{b}a)\\A=\arcsin(\frac{\sin B}{b}. a)\\A\approx 62.2^\circ$
Since the sum of the triangle is $180^\circ$, we obtain:
$$A+B+C=180^\circ\\C=180^\circ-A-B\\C=180^\circ -62.2^\circ -67.4^\circ\\C=50.4^\circ$$
