If tangents PA and PB from a point P to a cir | vedclass

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(C) Given that $PA$ and $PB$ are tangents to the circle from an external point $P$.The radius drawn to the point of contact is perpendicular to the ta

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(C) Given that $PA$ and $PB$ are tangents to the circle from an external point $P$.The radius drawn to the point of contact is perpendicular to the tangent.Therefore,$OA \perp PA$ and $OB \perp PB$,which implies $\angle OAP = 90^{\circ}$ and $\angle OBP = 90^{\circ}$.In the quadrilateral $OAPB$,the sum of interior angles is $360^{\circ}$.$\angle OAP + \angle APB + \angle OBP + \angle AOB = 360^{\circ}$$90^{\circ} + 80^{\circ} + 90^{\circ} + \angle AOB = 360^{\circ}$$260^{\circ} + \angle AOB = 360^{\circ}$$\angle AOB = 100^{\circ}$In $	riangle OPA$ and $	riangle OPB$:$OA = OB$ (Radii of the same circle)$PA = PB$ (Tangents from an external point are equal)$OP = OP$ (Common side)By $SSS$ congruence criterion,$	riangle OPA \cong 	riangle OPB$.By $CPCT$,$\angle POA = \angle POB$.Since $\angle POA + \angle POB = \angle AOB = 100^{\circ}$,we have $2 \angle POA = 100^{\circ}$.Therefore,$\angle POA = 50^{\circ}$.

If tangents $PA$ and $PB$ from a point $P$ to a circle with centre $O$ are inclined to each other at an angle of $80^{\circ}$,then $\angle POA$ is equal to ......$^{\circ}$

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