Question

If the 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

• A. ${50}^{\circ }$
• B. ${60}^{\circ }$
• C. ${70}^{\circ }$
• D. ${80}^{\circ }$

Answer 1

The correct option is A ${50}^{\circ }$

Given:$PA$ and $PB$ are tangents to circle and $\angle APB={80}^{\circ }$

Proof:Since $PA$ is tangent,$OA\perp PA$ (Tangent at any point of circle is perpendicular to the radius through point of contact)

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

In $△OAP$ and $△OBP$

$OA=OB$(radius of the circle)

$PA=PB$ (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

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

$\angle POA+{130}^{\circ }={180}^{\circ }$

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