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Question
In the given figure, AB is diameter of a circle centered at O. BC is tangent to the circle at B. If OP bisects the chord AD and ∠AOP = 60°, then find ∠C.
Solution
Since, OP bisects the chord AD, therefore ∠OPA = 90° ....[∵ The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord]
Now, In ΔAOP,
∠A = 180° – 60° – 90°
= 120° – 90°
= 30°
Also, we know that the tangent at any point of a circle is perpendicular to the radius through the point of contact
∴ ∠ABC = 90°
Now, In ΔABC,
∠C = 180° – ∠A – ∠B
= 180° – 30° – 90°
= 150° – 90°
= 60°
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