Question

In figure, two tangents PA and PB are drawn to a circle with Centre O from an external point P. Prove that angle APB equal to 2 angle OAB

Answer 1

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Answer 2

Heya !

Here's your answer

TO PROVE => angle AOB = 2(angle OAB)

Join OB

Now , we have , OA and OB as the radii of the circle .

So , angle OAB = angle OBA

In ∆OAB ,

angle AOB = 180° - 2(angle OAB)

Now , in quadrilateral OAPB ,

angle APB + angle OAP + angle OBP + angle AOB = 360° (angle sum property of quadrilateral)

=> As PA and PB are tangents ,

therefore ,

angle OAP = angle OBP = 90°

( radius of a circle is perpendicular to the point of contact of a tangent )

=> angle APB + 90° + 90° + angle AOB = 360°

=> angle APB + 180°-2(angle OAB) = 360°-180°

=> angle APB = 180°-180° + 2(angle OAB)

=> angle APB = 2(angle OAB)

Proved !!!!

Hope it helps!