WLOG $BC=6$. Construct $D$ on segment $BC$ with $BD=2$. The circle centered at $D$ thorugh $B$ intersects the perpendicular bisector of $BC$ twice, but since the two points are reflections of each otehr over $BC$, we only consider one of them. If $M$ is the midpoint of $BC$, then $DM=1$, $DO=2$, and $OM\perp DM$, so $OM=\sqrt{3}$, and thus $OB=2\sqrt{3}$. Now, $AD=2+2\sqrt{3},DC=4$, and $\angle ADC = 60^{\circ}$, so by LoC $AC=2\sqrt{6}$. Now we have $\angle BOC=120^{\circ},\angle AOC = 90^{\circ}$, so our answer is $\boxed{\angle A = 60^{\circ},\angle B = 45^{\circ}, \angle C = 75^{\circ}}$.