In figure, BC is a diameter of the circle and ∠BAO = 60° then, ∠ADC is equal to—

Given BC is a diameter of the circle and ∠BAO = 60°.

Consider ΔOAB,

⇒ OA = OB [Radii of same circle]

We know that angles opposite to equal sides are equal.

⇒∠OBA = ∠BAO = 60°

By angle sum property,

⇒∠OBA + ∠BAO + ∠AOB = 180°

⇒ 60° + 60° + ∠AOB = 180°

⇒∠AOB = 180° - 120°

∴ ∠AOB = 60°

By linear pair axiom,

⇒∠AOC = 180° - 60° = 120°

We know that angle subtended by an arc at the centre is double the angle subtended by the same arc at any point on the circle.

⇒∠AOC = 2 ∠ADC

⇒ 120° = 2 ∠ADC

⇒∠ADC = 120°/ 2

∴ ∠ADC = 60°