Q. 144.4( 7 Votes )

In the given figu

Answer :

In the given figure, PA and PB are two tangents from common point P


PA = PB


[ Tangents drawn from an external point are equal]


PBA = PAB


[ Angles opposite to equal angles are equal] …[1]


By angle sum property of triangle in APB


APB + PBA + PAB = 180°


60° + PAB + PAB = 180° [From 1]


2PAB = 120°


PAB = 60° …[2]


Now,


OAP = 90° [Tangents drawn at a point on circle is perpendicular to the radius through point of contact]


OAB + PAB = 90°


OAB + 60° = 90° [From 2]


OAB = 30°


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