Prove that the angle between 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.
This figure will surely help u to understand easily ......
To prove==> Angle BPA + Angle BOA = 180 ......
Solution==> In quadrilateral AOBP
Angle PBO = 900 [As angle between the radius and the tangent to the circle is 900]
Similarly angle PAO = 90.......
Angles (PAO + PBO + BOA + BPA) = 3600 ( sum of angles of a quadrilateral)
-----> 900 + 900 + angle BOA + angle BPA = 3600 (<PAO=90 and <PBO=90 proved)
-----> Angle BOA + Angle BPA = 3600 - 1800....
-----> Angle BOA + Angle BPA = 1800
!!!!....Hence proved.....Hope this helps....!!!!
