Answer
See below
Work Step by Step
We are given $A,a,c$. Use law of cosines to find $a$:
$$a^2=b^2+c^2-2bc\cos A\\b=\sqrt a^2+c^2-2ac\cos B\\b=\sqrt 8^2+6^2-2(8)(6)\cos 25^\circ\approx 3.6$$
Use law of sines to find: $\frac{\sin B}{b}=\frac{\sin C}{c}\\\sin C=\frac{\sin B}{b}\times c\\C=\arcsin(\frac{\sin B}{b}. c)\\C=\arcsin(\frac{\sin 25^\circ}{3.6}. 6)\approx50.8^\circ$
Since the sum of the triangle is $180^\circ$, we obtain:
$$A+B+C=180^\circ\\C=180^\circ-A-B\\A=180^\circ -25^\circ - 44.7^\circ\\A\approx 110.3 ^\circ$$
