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प्रश्न
In the given figure, O is the centre of circle. Find ∠AQB, given that PA and PB are tangents to the circle and ∠APB = 75°.
उत्तर
We know that
The radius is perpendicular to the tangent
∴ ∠OAP = ∠OBP = 90°
In quadrilateral PAOB
Sum of angles = 360°
∠APB + ∠OAP + ∠OBP + ∠AOB = 360°
75° + 90° + 90° + ∠AOB = 360°
255° + ∠AOB = 360°
∠AOB = 360° – 255°
∠AOB = 105°
Also, We know that
Angle subtended by an arc at the centre is double the angle subtended by the same arc at any point on the circle.
∴ ∠AOB = 2 × ∠AQB
105° = 2 × ∠AQB
2 × ∠AQB = 105°
∠AQB = `105^circ/2`
∠AQB = 52.5°
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