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

Given that PA and PB are tangents to circle.

∴ PA = PB. (∵ Lengths of tangents to circle from a fixed outer point are equal.)

∴ Triangle PAB is an isosceles triangle.

∴ ∠PAB = ∠PBA. (∵ Angles opposite to equal sides are always equal.)

∴ ∠PAB = 65°. (∵ ∠PBA = 65°(Given))

Since, PA is tangent to given circle and OA is radius of given circle at point of contact A.

∴ ∠PAO = 90°. (Angle between radius and tangent is 90° at point of contact)

And ∠PAO = ∠PAB + ∠OAB

⇒ ∠OAB = ∠PAO – ∠PAB = 90° –65° = 25°. (∵ ∠PAO = 90° & ∠PAB = 65°)

Now, in triangle PAB,

∠APB + ∠PAB + ∠PBA = 180°. (∵ Sum of angles in a triangle is 180°)

⇒ ∠APB = 180° – ∠PAB – ∠PBA = 180° – 65° – 65° = 180° – 130° = 150°.

(∵∠PAB = ∠PBA = 65°)

Hence, ∠OAB = 25° and ∠APB = 150°.