A geometry problem by Dinesh Grover

The sum of the interior angles are $180^o$ . So it can't exist 2 angles of $120^o$ because the sum of these angles are: $120^o+120^o=240^o>180^o$ Hence, the 2 other angles are: $\frac{180^o-120^o}{2}=30^o$ Therefore, the answer is $\boxed{30^o}$