
For convenience, take angle $BAO$ to be $\beta$, and $AO=OE=x$.
Now, we use the fact that the centroid divides the median into a ratio of $2:1$. Thus if $OE=x$ then $OB=2x$. And note that $\angle OBA= 180^{\circ}-(120^{\circ}+\beta)=60^{\circ}-\beta$. So using the sine rule on triangle $BOA$, we have:
$$\frac{\sin\beta}{2x}=\frac{\sin(60^{\circ}-\beta)}{x}$$
That gives us the equation, $\sin\beta=2\sin(60^{\circ}-\beta)$. Expanding, we get:
$$\sin \beta = 2\cos \beta\sin 60^{\circ}-2\cos 60^{\circ}\sin \beta$$
$$2\sin \beta=\sqrt{3} \cos \beta$$
$$\tan \beta= \frac{\sqrt{3}}{2}$$
Thus, $\beta=\arctan{\frac{\sqrt{3}}{2}}$. And hence, $\angle BAC =\arctan{\frac{\sqrt{3}}{2}}+60^{\circ}\approx 100.893^{\circ}$. Its worth noting here that though the solution is unique but is not a simple value what we would have hoped for, there is no convenient expression for the angle without inverse trig functions.