Question

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

1) ∠BCO

2) ∠AOB

3) ∠APB

Answer 1

In ΔAOC, ACO = 30° (Given)

∠OAC = 90° [radius is perpendicular to the tangent at the point of contact]
By angle sum property, ACO + OAC + AOC = 180o
AOC = 180° – (90° + 30°) = 60°
Consider Δ AOC and Δ BOC
AO = BO (radii)
AC = BC (tangents to a circle from an external point are equal in length)
OC = OC (Common)
ΔAOC ≅ ΔBOC

1) ∠BCO = ∠ACO = 30°

2) ∠AOC = ∠BOC = 60°

∠AOB = ∠AOC + ∠BOC = 120°

3) We know that, “If two angles stand on the same chord, then the angle at the centre is twice the angle at the circumference.

∠AOB and ∠APB stand on the same chord AB.

∠AOB = 2∠APB

So ∠APB = 1/2 ∠AOB = 60°