If from an external point P of a circle with centre O, two tangents PQ and PR are drawn such `/_QPR=120^@`, prove that 2PQ=PO.

Explanation

In `DeltaOPQ`, we have
`/_PQO=90^(@)` [`:'` the tangent at any point is perpendicular to the radius through the point of contact]
and `/_QPO=(1)/(2)xx/_QPR=(1)/(2)xx120^(@)=60^(@)`. [`:'` the two tangents drawn from an external point are equally inclined to the line segment joining the centre to that point and so `/_QPO=/_RPO`]
In right `DeltaOPQ`, we have
`cos(/_QPO)=(PQ)/(PO)`
`impliescos60^(@)=(PQ)/(PO)implies(1)/(2)=(PQ)/(PO)=2PQ=PO`.