Given : In circle C (O,r), OA = AB.
To find: ∠ADB and ∠ACB.
Solution: In ΔOAB
OA = AB [∵ Given]
OA = OB [∵ Radii of circle]
Hence, OA = OB = AB
⇒ ABC is an equilateral triangle.
Therefore, ∠AOB = 60° [∵ Each angle of an equilateral triangle is 60°]
∠AOB = 2 ∠ADB
[∵ The angle subtended by an arc at the centre is double the angle subtended by it at

any]
⇒ ∠ADB = 1/2 ∠AOB ⇒ ∠ADB = (1/2) × 60° = 30
ACBD is a cyclic quadrilateral.
Therefore, ∠ACB + ∠ADB = 180°
[∵ the sum of either pair of opposite angles of a cyclic quadrilateral is 180°]
⇒ ∠ACB + 30° = 180° [∵ ∠ADB = 30°]
⇒ ∠ACB = 180° – 30° = 150°