Equivalent to $OG$ parallel (or coincide with) to the $C$-bisector. If they coincide, then $ABC$ is isosceles and $\angle ACB = 100^{\circ}$. If not, then either do a trig bash or intersect the $C$-bisector with the circumcircle at $L$ and establish that $LO = 2OM$ where $OM \perp AB$, implying that $ABL$ is equilateral and hence $\angle ACB = 120^{\circ}$.

Coordinate bash also works very nicely here.

Turns out that pretty much the same problem appears in Bulgaria IMO 2012 TST.