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°

Answer :

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


[radii of same circle]


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°

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