Q. 305.0( 2 Votes )

In the given figure, PQR is a tangent to the circle at Q, whose center is O and AB is a chord parallel to PR such that BQR = 70°. Then, AQB =?


A. 20°

B. 35°

C. 40°

D. 45°

Answer :

In given figure, as PR is a tangent


OQ PR


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


LQ PR


LQ AB


[As, AB || PR]


AL = LB


[Perpendicular from center to the chord bisects the chord]


Now,


LQR = 90°


LQB + BQR = 90°


LQB + 70° = 90°


LQB = 20°…[1]


In AQL and BQL


ALQ = BLQ [Both 90° as LQ AB]


AL = LB [Proved above]


QL = QL [Common]


AQL BQL


[Side - Angle - Side Criterion]


LQA = LQB


[Corresponding parts of congruent triangles are congruent]


AQB = LQA + LQB = LQB + LQB


= 2LQB = 2(20) = 40° [By 1]

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