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°
