In figure 3, two tangents RQ and RP are drawn from an external point R to the circle with center O. If ∠PRQ = 120°, then prove that OR = PR + RQ.

Given : Two tangents RQ and RP from an external point R to the circle with center O and

PRQ = 120°

To Prove : OR = PR + QR

Construction : Join OP and OQ

Proof :

PR = RQ [ Tangents drawn from an external point to a circle are equal] …[1]

Also OP PR and OQ RQ

[ as tangent drawn at a point on a circle is perpendicular to the radius through the point of contact]

OPR = 90°

OQR = 90°

OPR + PRQ + OQR + QOP = 360°

90 + 120 + 90 + QOP = 360

QOP = 60°

In OPR and ORQ

OP = OQ [radii of same circle]

PR = RQ [By 1]

OR = OR [common]

OPR ORQ [By SAS]

POR = ROQ [By CPCT]

QOP = POR + ROQ

POR + POR = 60°

POR = 30°

As ROP is a right angled triangle

sin POR = PR/OR

sin 30° = PR/OR

1/2 = PR/OR

OR = 2PR

= PR + PR = PR + QR [from 1]

Hence Proved.

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