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If the angle between two tangents drawn from an external paint P to a circle of radius a and center O, is $60^{o}$, then find the length of OP.
Given: Radius of the circle $=a$ and angle between the two tangents drawn from external point $P=60^{o}$
To do: To find the length of OP.
Solution:
Given that $\angle BPA=60^{o}$
$OB=OA=a \ \ \ \ \ \ \ \ ( \ radii\ of\ the\ circle)$
$PA=PB\ \ \ \ \ \ [ length\ of\ tangents\ Equal\ OP=OP]$
$\vartriangle PBO$ and $\vartriangle PAO$ are congruent. [ By SSS congruency]
$\angle BPO=\angle OPA\ =\frac{60^{o}}{2} =30^{o}$
In $\vartriangle PBO$ ,
$sin\ 30^{o} =\frac{a}{OP}$
$\Rightarrow \frac{1}{2} =\frac{a}{OP}$
$\Rightarrow OP=2a\ units$
Updated on: 10-Oct-2022
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