Question

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

Answer 1

Since $PA$ is a tangent,$OP\perp PA$(Tangent at any point of circle is perpendicular to the radius through point of contact.)

$\therefore \angle OAP={90}^{\circ }$

In $△OAP$ and $△OBP$

$OA=OB$(both radius)

$PA=PB$(Lengths of tangent drawn from external point is equal)

$OP=OP$(common)

$\therefore △OAP\cong △OBP$ by SSS congruency

So,$\angle OPA=\angle OPB$ by CPCT

So,$\angle OPA=\frac{1}{2}\angle APB=\frac{1}{2}\times {80}^{\circ }={40}^{\circ }$

In $△OPA,$

$\angle POA+\angle OPA+\angle OAP={180}^{\circ }$ by angle sum property of triangle

$\angle POA+{40}^{\circ }+{90}^{\circ }={180}^{\circ }$

$\Rightarrow \angle POA={180}^{\circ }-{130}^{\circ }={50}^{\circ }$