In figure, if PA and PB are tangents to the circle with center O such that ∠APB = 50°, then ∠OAB is equal toA. 25°B. 30°C. 40°D. 50°

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]

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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