Question

In the given figure, PA and PB are two tangents from an external point P to a circle with centre O. If ∠PBA = 65^∘ , find ∠OAB and ∠APB

Answer 1

We know that tangents drawn from the external point are congruent.
∴ PA = PB
Now, In isoceles triangle APB
∠APB + ∠PBA + ∠PAB = 180^∘ [Angle sum property of a triangle]
⇒ ∠APB + 65^∘ + 65^∘ = 180^∘ [∵∠PBA = ∠PAB = 65^∘ ]
⇒ ∠APB = 50^∘
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^∘ + 50^∘ + 90^∘ = 360^∘
⇒ 230^∘ + ∠BOC = 360^∘
⇒ ∠AOB = 130^∘
Now, In isoceles triangle AOB
∠AOB + ∠OAB + ∠OBA = 180^∘ [Angle sum property of a triangle]
⇒ 130^∘ + 2∠OAB = 180⁰ [∵∠OAB = ∠OBA ]
⇒ ∠OAB = 25^∘