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)
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