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

Given: The angle between two tangents drawn from an external point $P$ to a circle of radius $r$ and a center $O$, is $60^o$

To do: To find the length of $OP$.

Solution:

$AP=BP$     ....$(length\ of\ tangents\ from\ external\ point\ to\ circle\ are\ equal)$

$\angle A=\angle B=90^o$  .... $( Tangent\ is\  \perp\ to\ radius)$ 

$OP=OP$        .... $( common\ side)$

$\because \vartriangle AOP\cong \vartriangle BOP$    ....$( RHS\ test\ of\ congruence)$

$\angle APO=\angle BPO=30^o$ $\rightarrow$c.a.c.t

$\angle AOP=\angle BOP=60^o$ $\rightarrow$c.a.c.t

$\vartriangle AOP$ is $30^o−60^o−90^o$ triangle.

In $\vartriangle AOP$, 

$cos60^o=\frac{OA}{OP}$
​
$\Rightarrow OP=\frac{r}{\frac{1}{2}}=2r$ 
 

Updated on: 10-Oct-2022

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