In the adjoining figure, O is the centre of the circle. Tangents to the circle at A and B meet at C. If ∠ACO = 30°, find

(i) ∠BCO

(ii) ∠AOB

(iii) ∠APB

(i) ∠BCO = ∠ACO = 30° (∵ C is the intersecting point of tangent AC and BC. So, OC divides ∠ACB in two halves.)

Hence, the value of ∠BCO = 30°.

(ii) We know that the tangent at any point of a circle and the radius through the point are perpendicular to each other.

∴ ∠OAC = ∠OBC = 90°.

∴ ∠AOC = ∠BOC (∵ tangents are equally inclined to the line joining the point and the centre of the circle.)

Since sum of angles in a triangle = 180.

In AOC

⇒ ∠AOC + ∠OAC + ∠ACO = 180°
⇒ ∠AOC + 90° + 30° = 180°
⇒ ∠AOC + 120° = 180°
⇒ ∠AOC = 180° - 120° = 60°.

∴ ∠BOC = 60°.

From figure,

∠AOB = ∠AOC + ∠BOC = 60° + 60° = 120°.

Hence, the value of ∠AOB = 120°.

(iii) Arc AB subtends ∠AOB at the centre and ∠APB at the remaining part of the circle.

∴ ∠AOB = 2∠APB (∵ angle subtended at centre by an arc is double the angle subtended at remaining point of the circle.)

∠APB = $\dfrac{1}{2}$ x ∠AOB = $\dfrac{1}{2}$ x 120° = 60°.

Hence, the value of ∠APB = 60°.