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Question
In the given figure, PA and PB are two tangents to the circle with centre O. If ∠APB = 50° then what is the measure of ∠OAB.
Solution
Construction: Join OB
We know that the radius and tangent are perpendicular at their point of contact
∴ ∠OBP = ∠OAP = 90°
Now, In quadrilateral AOBP
∠AOB + ∠OBP + ∠APB + ∠OAP = 360° [Angle sum property of a quadrilateral]
⇒ ∠AOB + 90° + 50° + 90° = 360°
⇒ 230° + ∠BOC= 360°
⇒ ∠AOB = 130°
Now, In isosceles triangle AOB
∠AOB + ∠OAB +∠OBA =180° [Angle sum property of a triangle]
⇒ 130° + 2∠OAB = 180° [∵ ∠OAB = ∠OAB]
⇒ ∠OAB = 25°
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