Q. 83.9( 8 Votes )

In figure, if PA and PB are tangents to the circle with center O such that APB = 50°, then OAB is equal to


A. 25°

B. 30°

C. 40°

D. 50°

Answer :

Given : A circle with center O and PA and PB are tangents to circle from a common external point P to point A and B respectively and APB = 50°


To find : OAB


OA AP and OB PB [ As tangent to at any point on the circle is perpendicular to the radius through point of contact]


OBP = OAP = 90° [1]


In Quadrilateral AOBP [ By angle sum property of quadrilateral]


OBP + OAP + AOB + APB = 360°


90° + 90° + AOB + 50° = 360°


AOB = 130° [2]


Now in OAB


OA = OB [Radii of same circle]


OBA = OAB [3]


Also, By angle sum property of triangle


OBA + OAB + AOB = 180°


OAB + OAB + 130 = 180 [using 2 and 3]


2OAB = 50°


OAB = 25°

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