Q. 165.0( 1 Vote )

If PA and PB are two tangents to a circle with center O such that AOB = 110° then APB is equal to


A. 55°

B. 60°

C. 70°

D. 90°

Answer :

As AP and BP are tangents to given circle,


We have,


OA AP and OB BP


[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]


So, OAP = OBP = 90°


In quadrilateral AOBP,


By angle sum property of quadrilateral, we have


OAP + OBP + AOB + APB = 360°


90° + 90° + 110° + APB = 360°


APB = 70°

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