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

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]

2∠OAB = 50°

∠OAB = 25°

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