Question

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°.

Answer 1

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°