AB is a diameter of a circle with centre O. CD is a chord equal to the radius of the circle. AC and BD are produced to meet at P. Then the measure of ∠APB is : ?->(Show Answer!)

1. AB is a diameter of a circle with centre O. CD is a chord equal to the radius of the circle. AC and BD are produced to meet at P. Then the measure of ∠APB is :

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By: anil on 05 May 2019 01.59 am

Given CD is equal to the radius.Thus triangle OCD is an equilateral triangle. ∴ ∠COD = 60°
Triangles OCA and triangles ODB are isosceles triangles as their two sides are radii.
In triangle OCA, OC = OA (both are radius)
∴ ∠OAC = ∠OCA (angles opposite to the equal sides are equal)
Let ∠OAC = ∠OCA = a
Thus ∠AOC = 180° - 2a
In triangle ODB, OD = OB (both are radius)
∴∠OBD = ∠ODB (angles opposite to the equal sides are equal)
Let ∠OBD = ∠ODB = b
Thus ∠BOD = 180° - 2b
Sum of angles in a straight line = 180°
∴At point O, (180° - 2a) + 60° + (180° - 2b) = 180°
2a + 2b = 240°
a + b = 120°
In triangle PAB , ∠APB + a + b = 180°
∠APB = 180° - a - b
∠APB = 180° - 120° = 60°

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