Q. 95.0( 1 Vote )

In the given figure, AB and AC are tangents to the circle with center O such that BAC = 40°. Then, BOC is equal to


A. 80°

B. 100°

C. 120°

D. 140°

Answer :

As AB and AC are tangents to given circle,


We have,


OB AB and OC AC


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


So, OBA = OCA = 90°


In quadrilateral AOBC, By angle sum property of quadrilateral, we have,


OBA + OCA + BOC + BAC = 360°


90° + 90° + BOC + 40° = 360°


BOC = 140°

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