Inscribed Quadrilateral Angle Measure

Since $\angle COD = 120^\circ$ is the angle subtended by chord $CD$ at the centre $O$, the angle subtended by $CD$ at the circumference is $\angle CAD = \frac{1}{2} \angle COD = 60^\circ$.

Now, $\angle BAD = \angle BAC + \angle CAD = 30^\circ + 60^\circ = 90^\circ$.

Since $ABCD$ is a cyclic quadrilateral, opposite angles add up to $180^\circ$: $\angle BCD = 180^\circ - \angle BAD = 180^\circ - 90^\circ = 90^\circ$.