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If from an external point of a circle with center O, two tangents PQ and PR are drawn such that $\angle QPR= 120^{o}$, prove that $2PQ=PO$.
Given: An external point of a circle with center O, two tangents PQ and PR are drawn such that $\angle $QPR = 120^{o}$
To do: To prove that $2PQ=PO$.
Solution:
Let us draw the circle with extent point P and two tangents PQ and PR.
We know that the radius is perpendicular to the tangent at the point of contact .
$\angle OQP=90^{o} $
We also know that the tangents drawn to a circle from an external point are equally inclined to the joining the center to that point.
$\angle QPO=60^{o}$
Now, in $\vartriangle QPO$:
$cos60^{o}=\frac{PQ}{PO}$
$\Rightarrow \frac{1}{2} =\frac{PQ}{PO}$
$\Rightarrow PO=2PQ$
Hence proved that $PO=2PQ$.
Updated on: 10-Oct-2022
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