Question

In the figure PA and PB are tangents to the circle. If $\angle$ APO $={30}^{\circ }$ , find $\angle$ AOB.

Answer 1

$\angle$ APO $={30}^{\circ }$ .....given
From P we have two tangents PA and PB
We know that if we join point P and center of circle
O then the line PO divides the angle between tangents
$\Rightarrow \angle$ APO = $\angle$ OPB = ${30}^{\circ }$ .....(i)
$\angle$ OAP = $\angle$ OBP = ${90}^{\circ }$ ....... radius is perpendicular to tangent ....(ii)
Consider quadrilateral OAPB
$\Rightarrow \angle$ OAP + $\angle$ APB + $\angle$ PBO + $\angle$ AOB = ${360}^{\circ }$ ....sum of angles of quadrilateral
From figure $\angle$ APB = $\angle$ APO + $\angle$ OPB
$\Rightarrow \angle$ OAP + $\angle$ APO + $\angle$ OPB + $\angle$ PBO + $\angle$ AOB = ${360}^{\circ }$
Using (i) and (ii)
$\Rightarrow {90}^{\circ }+{30}^{\circ }+{30}^{\circ }+{90}^{\circ }+\angle$ AOB = ${360}^{\circ }$
$\Rightarrow {240}^{\circ }+\angle$ AOB = ${360}^{\circ }$
$\Rightarrow \angle$ AOB = ${120}^{\circ }$
Hence $\angle$ AOB is ${120}^{\circ }$