#### Question

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

(i) ∠BCO (ii) ∠AOB (iii) ∠APB

#### Solution

In the given fig, O is the centre of the circle and CA and CB are the tangents to the circle from

C. Also, ∠ACO = 30°

P is any point on the circle. P and PB are joined.

To find: (i) BCO

(ii) `∠`AOB

(iii)`∠`APB

Proof:

(i) In ΔOAC and OBC

OC = OC (Common)

OA = OB (radius of the circle)

CA = CB (tangents to the circle)

∴ ΔOAC ≅ ΔOBC (SSS congruence criterion)

∴ `∠`ACO = `∠`BCO = 30°

(ii) ∴ `∠`ACB = 30° + 30° = 60°

∴ `∠`AOB + `∠`ACB = 180°

⇒ `∠`AOB + 60° = 180°

⇒ `∠`AOB = 180° - 60°

⇒ `∠`AOB = 120°

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

∴ `∠APB = 1/2 ∠AOB =1/2 xx120 = 60°`