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AB is the diameter of a circle, centre O. C is a point on the circumference such that ∠COB = θ. The area of the minor segment cutoff by AC is equal to twice the area of sector BOC. Prove that sinθ/2 .cosθ/2=π(1/2−θ/120°)

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Given AB is diameter of circle with centre O

∠COB = θ

Area of segment cut off, by AC = (area of sector) – (area of ΔAOC)

∠AOC = 180 – θ [∠AOC and ∠BOC form linear pair]

In ΔAOC, drop a perpendicular AM, this bisects ∠AOC and side AC.

Now, In ΔAMO,

Area of segment by AC = 2 (Area of sector BDC)

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