Question

Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

Answer 1

To Prove: ∠APB + ∠AOB = 180° 

OA is radius, PA is tangent. 

∴∠PAO = 90° OB is radius, PB is tangent. 

∴ ∠PBO = 90° 

Now OAPB is a quadrilateral. 

∴∠PAO + ∠PBO = 90° + 90° = 180° 

Sum of four angles of a quadrilateral is 360° 

∴ ∠PAO + ∠PBO + ∠APB + ∠AOB = 360° 

180° + ∠APB + ∠AOB = 360° 

∠APB + ∠AOB = 360° – 180° 

∴ ∠APB + ∠AOB = 180° 

If sum of two angles is equal to 180°, then they are supplementary angles. 

∴ ∠APB and ∠AOB are supplementary angles.