Question

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

Answer 1

OA and OB are radii of the circle to the tangents PA and PB respectively.
The line drawn from the centre of the circle to the tangent is perpendicular to the tangent.
$\therefore OA\perp PA$ and $OB\perp PB\angle OBP=\angle OAP={90}^{\circ }$
In quadrilateral AOBP,
Sum of all interior angles = ${360}^{\circ }$
$\angle AOB+\angle OBP+\angle OAP+\angle APB={360}^{\circ }$


$\Rightarrow \angle AOB+{90}^{\circ }+{90}^{\circ }+{80}^{\circ }={360}^{\circ }\Rightarrow \angle AOB={100}^{\circ }$
Now,
In $\Delta$ OPB and $\Delta$ OPA,
AP = BP (Tangents from a point are equal)
OA = OB (Radii of the circle)
OP = OP (Common side)
$\therefore OPB\cong \Delta OPA$ (by SSS congruence condition)
Thus, $\angle POB=\angle POA\angle AOB=\angle POB+\angle POA\Rightarrow 2\angle POA=\angle AOB\Rightarrow \angle POA={100}^{\circ }/2={50}^{\circ }$
$\angle POA={100}^{\circ }/2={50}^{\circ }$