Question

In the given circle with centre O, PA and PB are tangents and ∠OAB = 28°, then ∠APB is:

Options

• 90°

• 56°

• 62°

• 90° + 28°

Answer 1

56°

Explanation:

We know that,

Tangent at any point of a circle and the radius through this point are perpendicular to each other.

∴ ∠OAP = 90°

From figure,

⇒ ∠PAB = ∠OAP − ∠OAB

= 90° − 28°

= 62°

PA and PB are tangents drawn to the circle from the external point P.

∴ PA = PB

We know that,

Angles opposite to equal sides of a triangle are equal.

∴ ∠PAB = ∠PBA = 62°

By angle sum property of triangle PAB,

⇒ ∠PAB + ∠PBA + ∠APB = 180°

⇒ 62° + 62° + ∠APB = 180°

⇒ ∠APB = 180° − 124°

= 56°