Answer
See below
Work Step by Step
We are given $a, b,c$. Use law of cosines to find $A$:
$$a^2=b^2+c^2-2bc\cos A\\ 18^2=28^2+13^2-2(28)(13)\cos A\\\cos A=\frac{629}{728}\\A=\cos^{-1}(\frac{629}{728})=30.2 ^\circ$$
Use law of sines to find: $\frac{\sin B}{b}=\frac{\sin A}{a}\\\sin B=\frac{\sin A}{a}\times b\\\arcsin (\sin B)=\arcsin (\frac{\sin A}{a}b)\\B=\arcsin(\frac{\sin A}{a}. b)\\B=\arcsin(\frac{\sin 30.2^\circ}{18}. 28)\approx 51.6^\circ$
OR $B \approx 128.4^\circ$
Since the sum of the triangle is $180^\circ$, we obtain:
$$A+B+C=180^\circ\\C=180^\circ-A-B\\C=180^\circ -30.2^\circ - 51.6
^\circ\\C\approx 98.2^\circ$$
For $B=128.4 ^\circ$, we have $C=180^\circ -30.2^\circ - 128.4
^\circ=21.4^\circ$
