In a cyclic quadrilateral ABCD ∠BCD=120° and AB passes through the centre of the circle. Then ∠ABD = ?

Given : ABCD is a cyclic quadrilateral and ∠BCD=120°

To find : ∠ABD = $$\theta$$ = ?

Solution : Sum of opposite angles of a cyclic quadrilateral = $$180^\circ$$

=> $$\angle$$ BCD + $$\angle$$ BAD = $$180^\circ$$

=> $$\angle$$ BAD = $$180-120=60^\circ$$

Also, the angle subtended by an arc at the centre is double the angle subtended by it at any point on the circle.

=> $$\angle$$ ADB = $$\frac{\angle AOB}{2}=\frac{180}{2}=90^\circ$$

Now, in $$\triangle$$ ABD,

=> $$\angle$$ BAD + $$\angle$$ ADB + $$\angle$$ ABD = $$180^\circ$$

=> $$60^\circ+90^\circ+\theta=180^\circ$$

=> $$\theta=180-150=30^\circ$$

=> Ans - (A)