Answer
See below
Work Step by Step
We are given $A, b,c$. Use law of cosines to find $a$:
$$a^2=b^2+c^2-2bc\cos A\\a=\sqrt b^2+c^2-2bc\cos A\\a=\sqrt 23^2+26^2-2(23)(26)\cos 114^\circ\approx 41.13$$
Use law of sines to find: $\frac{\sin B}{b}=\frac{\sin A}{a}\\\sin B=\frac{\sin A}{a}\times b\\\arcsin (\sin B)=\arcsin (\frac{\sin A}{a}b)\\B=\arcsin(\frac{\sin A}{a}. b)\\B=\arcsin(\frac{\sin 114^\circ}{41.13}. 23)\approx 30.7^\circ$
Since the sum of the triangle is $180^\circ$, we obtain:
$$A+B+C=180^\circ\\C=180^\circ-A-B\\C=180^\circ -114^\circ - 30.7^\circ\\C\approx 35.3 ^\circ$$
