Question

If the angle between two tangents drawn from an external point $P$ to a circle of radius $a$ and centre $O$, is ${60}^o$, then find the length of $OP$.

Answer 1

Given: $\angle APB={60}^{\circ }$
Consider $\Delta OPA$ and $\Delta OPB$ then
$\angle OAP=\angle OBP$ [ Angles rights angled]
$OP=OP$ [Common Side]
$OA=OB$ [Radius]
Therefore, $\Delta OPA\cong \Delta OPB$
So, $\angle OPA=\angle OPB=\frac{\angle BPA}{2}$
$\therefore \angle OPA=\frac{60}{2}={30}^{\circ }$
Thus, from right angled triangle $\Delta APO$, we have
$\sin {30}^{\circ }=\frac{AO}{PO}$
$\Rightarrow \frac{1}{2}=\frac{a}{OP}$
$\therefore OP=2a$