Question

A tangent ST to a circle touches it at B. AB is a chord such that ∠ABT = 65°. Find ∠AOB, where “O” is the centre of the circle.

Answer 1

Given ∠ABT = 65°

∠OBT = 90°  ...(TB is the tangent of the circle)

∠ABO = 90° – 65° = 25°

∠ABO + ∠BOA + ∠OAB = 180°

25° + x + 25° = 180° ...(Sum of the angles of a ∆)

OA and OB are the radius of the circle.

∴ ∠ABO = ∠BAO = 25°

x + 50 = 180°

x = 180° – 50° = 130°

∴ ∠BOA = 130°