Q4 of 14 Page 1

In Figure 2, AB and AC are tangents to the circle with centre O such that ∠BAC = 40ᵒ. Then ∠BOC is equal to


We know tangent drawn at a point on the circle is perpendicular to the radius through point of contact, so we have


∠BAC = 40° [Given]


∠ABO = 90°


∠ACO = 90°


Now, By angle sum property of quadrilateral, we have


⇒ ∠BAC + ∠ABO + ∠BOC + ∠OCA = 360°


⇒ ∠BOC = 360° – 40° – 90° – 90°


⇒ ∠BOC = 140°


Hence option C is correct.

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