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.

Answer 1

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°