AB is a diameter of a circle with centre O, CD is a chord equal t

SOLUTION

Given CD is equal to the radius.

Thus triangle OCD is an equilateral triangle.

∴ ∠COD = 60°

Triangles OCA and triangles ODB are isosceles triangles as their two sides are radii.

In triangle OCA,

OC = OA (both are radius)

∴ ∠OAC = ∠OCA (angles opposite to the equal sides are equal)

Let ∠OAC = ∠OCA = a

Thus ∠AOC = 180° - 2a

In triangle ODB,

OD = OB (both are radius)

∴∠OBD = ∠ODB (angles opposite to the equal sides are equal)

Let ∠OBD = ∠ODB = b

Thus ∠BOD = 180° - 2b

Sum of angles in a straight line = 180°

∴At point O,

(180° - 2a) + 60° + (180° - 2b) = 180°

⇒ 2a + 2b = 240°

⇒ a + b = 120°

In triangle PAB

∠APB + a + b = 180°

⇒ ∠APB = 180° - a – b

⇒ ∠APB = 180° - 120° = 60°