Question

In tangents $PA$ and $PB$ from a point $P$ to a circle with center $O$ are inclined to each other at an angle of ${80}^{\circ }$, then find $\angle POA$ (in degrees).

Answer 1

As it is given that PA and PB are tangents.
Therefore, the radius drawn to these tangents will be perpendicular to the tangents.
Thus, OA $\perp$ PA and OB $\perp$ PB
$\angle OBP$ = ${90}^{\circ }$ and $\angle OAP$ = ${90}^{\circ }$
In $△AOBP$, Sum of all interior angles =${360}^{\circ }$
$\angle OAP$ + $\angle APB$ +$\angle PBO$ + $\angle BOA$ = ${360}^{\circ }$
${90}^{\circ }$ + ${80}^{\circ }$ + ${90}^{\circ }$ + $\angle BOA$ =${360}^{\circ }$
$\angle BOA$ = ${90}^{\circ }$
In $\angle OPB$ and $\angle OPA$,
AP = BP (Tangents from a point)
OA = OB (Radii of the circle)
OP = OP (Common side)
Therefore, $△OPB$ $\cong$ $△OPA$ (SSS congruence criterion)
And thus, $\angle POB$
$\angle POA$ = $\frac{1}{2}$ $\angle AOB$ $=$ $\frac{{100}^{\circ }}{2}$ = ${50}^{\circ }$