Question

PA and PB are tangents drawn to the circle with centre O as shown in the figure. Prove that ∠APB = 2∠OAB.

Two tangents PA and PB are drawn to a circle with centre O from an external point P. Prove that ∠APB = 2∠OAB.

Answer 1

PA = PB   ...(Tangents from an external point to a circle)

∠PAB = ∠PBA = x   ...(angles opposite to equal sides) 

In ∆PAB,

∠PAB + ∠PBA + ∠APB = 180° 

x + x +∠APB = 180° 

∠APB = 180° − 2x   ...(i) 

Also, 

∠PAB + ∠OAB = 90°   ...(radius is perpendicular to the tangent at the point of contact) 

x +∠OAB = 90°

x = 90° − ∠OAB   ...(ii) 

Substituting (ii) in (i), we get

∠APB = 180° − 2(90° − ∠OAB) 

∠APB = 2∠OAB