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.

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#### 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°

Concept: Concept of Circle - Centre, Radius, Diameter, Arc, Sector, Chord, Segment, Semicircle, Circumference, Interior and Exterior, Concentric Circles

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