Class 10

Math

All topics

Conic Sections

If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of $80_{∘}$, then $∠POA$ is equal to

Please refer to the video for the diagram.

We know tangent at any point $P$ is always perpendicular to line joining the center $O$ of that circle.

So, $∠OAP=∠OBP=90_{∘}$

Now, in $ΔOAP$ and $ΔOBP$,

$OA=OB,OP=OP,∠OAP=∠OBP=90_{∘}$

So, $ΔOAPΔOBP$.

It means, $∠APO=∠BPO$

We are given, $∠APB=80_{∘}$

So, $∠APO=∠BPO=280 =40_{∘}$

In $ΔAOP$,

$∠AOP+∠APO+∠OAP=180_{∘}$

$∠AOP=180−40−90=50_{∘}$