Question

From an external point P, tangents PA and PB are drawn to a circle with centre O. If ∠PAB = 50°, then find ∠AOB.

Answer 1

It is given that PA and PB are tangents to the given circle.

∴∠PAO=90°      (Radius is perpendicular to the tangent at the point of contact.)

Now

∠PAB=50°          (Given)

∴∠OAB=∠PAO−∠PAB=90°−50°=40°

In ∆OAB,

OB = OA    (Radii of the circle)

∴∠OAB=∠OBA=40°           (Angles opposite to equal sides are equal.)

Now

∠AOB+∠OAB+∠OBA=180°       (Angle sum property)

⇒∠AOB=180°−40°−40°=100°