Question

Two tangents $PA$ and $PB$ are drawn to a circle with centre $O$. Such that $\angle APB={120}^{\circ }$. Prove that $OP=2AP$.

Answer 1

In $△OAP\&△OBP$
$OP=OP$(common)
$\angle OAP=\angle OBP\left(90°\right)$($\because$ Radius is $\perp$ to the tangents at the point of contact.)
$OA=OB$ (Radius of circle)
$\therefore OAP\cong △OBP(\because$ RUS)
$\angle OPA=\angle OPB=\frac{120}{2}=60$(CPCT)
In $△OAP,\cos \angle OAP=\cos 60=\frac{AP}{OP}\frac{1}{2}=\frac{AP}{OP}OP=2AP$
Hence proved.