Q. 24

If PA and PB are two tangents to a circle with center O such that ∠APB = 80°. Then, ∠AOP = ?A. 40°B. 50°C. 60°D. 70°

In Given Figure,

PA = PB…[1]

[Tangents drawn from an external point are equal]

In AOP and BOP

PA = PB [By 1]

OP = OP [Common]

OA = OB

AOP ≅△BOP

[By Side - Side - Side Criterion]

OPA = OPB

[Corresponding parts of congruent triangles are congruent]

Now,

APB = 80° [Given]

OPA + OPB = 80°

OPA + OPA = 80°

2 OPA = 80°

OPA = 40°

In AOP,

OAP = 90°

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

And

OAP + OPA + AOP = 180°

90° + 40° + AOP = 180°

AOP = 50°

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 Circles37 mins
Short Cut Trick to Find Area of Triangle43 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