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