Question

In figure, PA and PB are tangents to the circle with centre O such that $\angle APB={50}^{\circ }$ Write the measure of $\angle OAB$

Answer 1

Given PA & PB are tangent to the circle with center O.

$PA=PB$ [length of tangent from external point to circle are equal]

In $\Delta PAB$

$PA=PB$

$\angle PBA=\angle PAB$ [isosceles triangle]

now $\angle PAB+\angle PBA+\angle APB={180}^o$ [Angle sum prop]

$2\angle PAB=180-50=130$

$\angle PBA=\angle PAB={65}^o$ ………..$\left(1\right)$

Now PA is tangent & OA is radius at point A.

$\angle OAP={90}^o$ [tangent at any point is $\perp$ to radius]

$\angle OAB=\angle OAP-\angle PAB=90-65={25}^o$

Hence angle OAB is ${25}^o$.