Q. 54.4( 8 Votes )

In figure 3, two

Answer :


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°


In quadrilateral PORQ


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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