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From an external point P, tangents PA and PB are drawn to a circle with centre O. If ∠PAB = 50°, then find ∠AOB.

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#### Solution

It is given that PA and PB are tangents to the given circle.

∴∠PAO=90° (Radius is perpendicular to the tangent at the point of contact.)

Now

∠PAB=50° (Given)

∴∠OAB=∠PAO−∠PAB=90°−50°=40°

In ∆OAB,

OB = OA (Radii of the circle)

∴∠OAB=∠OBA=40° (Angles opposite to equal sides are equal.)

Now

∠AOB+∠OAB+∠OBA=180° (Angle sum property)

⇒∠AOB=180°−40°−40°=100°

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