Question

In the given figure, PA and PB are two tangents to the circle with centre O. If ∠APB = 60^∘ then ∠OAB is [CBSE 2011]
(a) 15^∘
(b) 30^∘
(c) 60^∘
(d) 90^∘

Answer 1

Construction: Join OB


We know that the radius and tangent are perperpendular at their point of contact
∵∠OBP = ∠OAP = 90^∘
Now, In quadrilateral AOBP
∠AOB + ∠OBP + ∠APB + ∠OAP = 360^∘ [Angle sum property of a quadrilateral]
⇒ ∠AOB + 90^∘ + 60^∘ + 90^∘ = 360^∘
⇒ 240^∘ + ∠AOB = 360^∘
⇒ ∠AOB = 120^∘
Now, In isoceles triangle AOB
∠AOB + ∠OAB + ∠OBA = 180^∘ [Angle sum property of a triangle]
⇒ 120^∘ + 2∠OAB = 180^∘ [∵∠OAB = ∠OBA ]
⇒ ∠OAB = 30^∘
Hence, the correct answer is option (b).