Listen NCERT Audio Books to boost your productivity and retention power by 2X.
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.
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 = 2∠TRQ
110° = 2 ∠TRQ
∠TRQ = 55°