Answer
$A=73.8^{\circ}$; $B=53.1^{\circ}$ and $C=53.1^{\circ}$
Work Step by Step
Need to apply the law of cosines to find $B$ when the sides of a triangle are given.
$b^2 = a^2+c^2-2ac \ cos \ B$
This implies that $B = arccos(\dfrac{a^2+c^2-b^2}{2ac}) = arccos(\dfrac{12^2+10^2-10^2}{(2)(12)(10)})= arccos(0.6)= 53.1^{\circ}$
Since, by symmetry, $ \angle C = 53.1^{\circ}$
Now, find angle $A$.
$A+B+C = 180^{\circ} $
or, $A= 180^{\circ}-B-C\\= 180^{\circ}-53.1^{\circ}-53.1^{\circ}\\
= 73.8^{\circ}$
