Question

From an external point P , tangents PA = PB are drawn to a circle with centre O   . If  \[\angle PAB = {50}^o\] , then find  \[\angle AOB\]

Answer 1

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

\[\therefore \angle PAO = 90^o\]   (Radius is perpendicular to the tangent at the point of contact.)

Now,

\[\angle PAB = 50^o\]          (Given)

\[\therefore \angle OAB = \angle PAO - \angle PAB = 90^o - 50^o = 40^o\]

In ∆OAB,
OB = OA    (Radii of the circle)

\[\therefore \angle OAB = \angle OBA = 40^o\] (Angles opposite to equal sides are equal.)

Now,

\[\angle AOB + \angle OAB + \angle OBA = 180^o\] (Angle sum property) 

\[\Rightarrow \angle AOB = 180^o - 40^o - 40^o= 100^o\]