Question

If tangents $PA$ and $PB$ from a point to a circle with centre $O$ are inclined to each other at an angle of ${80}^o$, then $\angle POA$ is equal to

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

Answer 1

Answer: (A)

${50}^o$

Given-

O is the center of a circle to which two tangents $PA\&PB$ have been drawn at $A\&B$ respectively.

$\angle APB={80}^o$.

To find out-$\angle POA$

Solution-

We join $OA\&OB\&OP$.

So, OA & OB are radii through the points of contact of the tangents PA & PB to the given circle respectively.
$\therefore OA\perp PA$ & $OB\perp PB$ since the radius through the point of contact of a tangent to a circle is perpendicular to the tangent.
$\therefore \angle OAP={90}^o=\angle OBP$.

Now, in quadrilateral PAOB we have $\angle AOB={360}^o-\angle OAP-\angle OBP-\angle APB$

$={360}^o-{90}^o-{90}^o-{80}^o={100}^o.$

Now, between $\Delta OPA$ & $\Delta OPB$ we have

$OA=OB$ ...(radii of the same circle),

$PA=PB$ ...(Tangents to the same circle from a single point are equal) and OP is common side.
$\Delta OPA\cong \Delta OPB$ ....by $SSS$ test,

$⟹\angle POA=\angle POB=\frac{1}{2}\angle AOB=\frac{1}{2}\times {100}^o={50}^o$