# 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°

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]

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Quiz | Imp. Qs. on Circle35 mins
Quiz | Imp. Qs. on Circles37 mins
Quiz | Testing Your Knowledge on Circles32 mins
Short Cut Trick to Find Area of Triangle43 mins
Quiz | Areas Related to Circles43 mins
RD Sharma | Area of Sector and Segments25 mins
Quiz | Area Related with Circles47 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses