# In figure, if PQR is the tangent to a circle at Q whose center is 0, AB is a chord parallel to PR and ∠BQR = 70°, then ∠AQB is equal toA. 20°B. 40°C. 35°D. 45°

Given : PQR is a tangent to a circle Q with center O and AB is a chord parallel to PR and BQR = 70°

To Find : AQB

Now, OQR = DQR = 90° [ As tangent at any point on the circle is perpendicular to the radius through point of contact]

DQR = QBD + BQR

90 = QBD + 70°

QBD = 20° [1]

As, AB || PR

BDQ + DQR = 180° [ some of co interior angles ]

BDQ + 90° = 180°

BDQ = 90°

And

ADQ + BDQ = 180° [Linear pair]

In AQD and BQD

QD = QD [common]

AD = BD [since, perpendicular from center to the chord bisects the chord]

AQD BQD [By Side Angle Side Criterion]

And

AQD = BQD [ Corresponding parts of congruent triangles are equal ]

AQD + BQD = AQB

AQB = 2BQD [As AQD = BQD ]

AQB = 2(20) = 40° [From 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 | Testing Your Knowledge on Circles32 mins
Important Questions on Circles46 mins
Quiz | Imp. Qs. on Circle35 mins
Smart Revision | Circles44 mins
Quiz | Imp. Qs. on Circles37 mins
RD Sharma | Important Questions on Circles36 mins
Short Cut Trick to Find Area of Triangle43 mins
Tangent from an External Point54 mins
RD Sharma | Most Important Questions of Circles35 mins
RD Sharma | Imp Qs Discussion on Area Related With Circles41 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