Given : ABCD is a cyclic quadrilateral and ∠BCD=120° To find : ∠ABD = $$ heta$$ = ? 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 $$ riangle$$ ABD, => $$angle$$ BAD + $$angle$$ ADB + $$angle$$ ABD = $$180^circ$$ => $$60^circ+90^circ+ heta=180^circ$$ => $$ heta=180-150=30^circ$$ => Ans - (A)