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In the given figure, ∠BOQ = 60° and AB is diameter of the circle. Find ∠ABO.

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A20°

B30°

C40°

D50°

Answer:

B. 30°

Read Explanation:

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Using the theorem, angle in the semi-circle is a right angle,

⇒ ∠BOA = 90°

Theorem: The alternate segment theorem states that an angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment.

⇒ ∠BOQ = ∠BAO = 60°

In ΔABO,

Sum of the angles of the triangle is 180°

⇒ ∠ABO = 180° – ∠BOA – ∠BAO = 180° – 90° – 60° = 30°


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