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