Q. 34.2( 271 Votes )

If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then POA is equal to

A. 50° B. 60°

C. 70° D. 80°

Answer :

Given: PA and PB are tangents.



Therefore, the radius drawn to these tangents will be perpendicular to the tangents.


Thus, OA PA and OB PB


OBP = 90°


OAP = 90°


In quadrilateral AOBP,


Sum of all interior angles = 360°


OAP + APB +PBO + BOA = 360° 90° + 80° +90° + BOA = 360°


BOA = 100°


In ΔOPB and ΔOPA,


AP = BP (Tangents from a point)


OA = OB (Radii of the circle)


OP = OP (Common side)


Therefore, ΔOPB ΔOPA (SSS congruence criterion)


And thus, POB = POA


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