Answer 1

Let us draw the circle with external point P and two tangents PQ and PR.

We know that the radius is perpendicular to the tangent at the point of contact.

∴∠OQP=90°

We also know that the tangents drawn to a circle from an external point are equally inclined to the segment, joining the centre to that point.

∴∠QPO=60°

Now in ∆QPO:

`cos60^@="PQ"/"PO"`

`⇒1/2="PQ"/"PO"`

`⇒2PQ=PO`