Question

Two tangents PA and PB are drawn from an external point P to the circle with centre O, such that $\angle$APB =${120}^{\circ }$. What is the relation between OP and AP?

• A. $OP=\frac{1}{2}\times AP$
• B. $AP=\frac{2}{3}\times OP$
• C. OP = 2(AP)
• D. OP = AP

Answer 1

The correct option is C OP = 2(AP)

Consider $△APOand△BPO$,
AO = BO [Radius of the circle]
AP = BP [Tangents from the same external point]
OP = OP [Common side]

$\therefore$ $△APO\cong △BPO$
[By SSS congruency]
$⟹\angle OPA=\angle OPB$
[By CPCT]
Given that $\angle APB={120}^{\circ }$
$\therefore \angle OPA+\angle OPB=2\angle OPA={120}^{\circ }$
So, $\angle OPA={60}^{\circ }$.

We know that $△OPA$ is a right angled triangle with $\angle OAP={90}^{\circ }$.

$Thus,cos\angle OPA=\frac{AP}{OP}=cos{60}^{\circ }$
$⟹\frac{AP}{OP}=\frac{1}{2}$

$Thus,OP=2AP$.