Q. 35.0( 8 Votes )

In the given figure, O is the center of a circle. PT and PQ are tangents to the circle from an external point P. If TPQ = 70°, find TRQ.


Answer :


Given: In the figure, PT and PQ are two tangents and TPQ = 70°


To Find: TRQ


Construction: Join OT and OQ


In quadrilateral OTPQ


OTP = 90°


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


OQP = 90°


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


TPQ = 70° [Common]


By Angle sum of Quadrilaterals,


In quadrilateral OTPQ we have


OTP + OQP + TPQ + TOQ = 360°


90° + 90° + 70° + TOQ = 360°


250° + TOQ = 360


TQO = 110°


Now,


As we Know the angle subtended by an arc at the center is double the angle subtended by it at any


point on the remaining part of the circle.


we have


TOQ = 2TRQ


110° = 2 TRQ


TRQ = 55°


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