In the figure,if PA and PB are tangents to th | vedclass

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(C) Given that $PA$ and $PB$ are tangents to the circle from an external point $P$.Since the lengths of tangents drawn from an external point to a cir

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$25$

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(C) Given that $PA$ and $PB$ are tangents to the circle from an external point $P$.Since the lengths of tangents drawn from an external point to a circle are equal,we have $PA = PB$.In $	riangle PAB$,since $PA = PB$,the angles opposite to these sides are equal,i.e.,$\angle PBA = \angle PAB$.Let $\angle PBA = \angle PAB = 	heta$.In $	riangle PAB$,the sum of angles is $180^{\circ}$,so $\angle APB + \angle PAB + \angle PBA = 180^{\circ}$.Substituting the given values: $50^{\circ} + 	heta + 	heta = 180^{\circ}$.$2	heta = 180^{\circ} - 50^{\circ} = 130^{\circ}$.$	heta = 65^{\circ}$.We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact. Therefore,$OA \perp PA$,which means $\angle OAP = 90^{\circ}$.Since $\angle OAP = \angle OAB + \angle PAB$,we have $90^{\circ} = \angle OAB + 65^{\circ}$.$\angle OAB = 90^{\circ} - 65^{\circ} = 25^{\circ}$.

In the figure,if $PA$ and $PB$ are tangents to the circle with center $O$ such that $\angle APB = 50^{\circ}$,then $\angle OAB$ is equal to: (in $^{\circ}$)

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