overline{ PA } and overline{ PB } are the tan | vedclass

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(D) In the quadrilateral $OAPB$,the radius is perpendicular to the tangent at the point of contact.Therefore,$\angle OAP = 90^{\circ}$ and $\angle OBP

Vedclass · Accepted Answer

$115$

Vedclass · Answer

(D) In the quadrilateral $OAPB$,the radius is perpendicular to the tangent at the point of contact.Therefore,$\angle OAP = 90^{\circ}$ and $\angle OBP = 90^{\circ}$.The sum of the angles in a quadrilateral is $360^{\circ}$.Thus,$\angle AOB + \angle OAP + \angle APB + \angle OBP = 360^{\circ}$.Substituting the known values: $\angle AOB + 90^{\circ} + 65^{\circ} + 90^{\circ} = 360^{\circ}$.$\angle AOB + 245^{\circ} = 360^{\circ}$.$\angle AOB = 360^{\circ} - 245^{\circ} = 115^{\circ}$.

$\overline{ PA }$ and $\overline{ PB }$ are the tangents to $\odot( O , r)$ drawn from a point $P$ outside a circle. If $m \angle APB = 65^{\circ}$,then $m \angle AOB = \ldots \ldots \ldots . .$ (in $^{\circ}$)

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