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AB is the diameter of a circle, center O.C is a point on circumference such that angle COB=θ. The area of the minor segment cut off by AC is equal to twice the area of sector BOC. Prove that
sinθ2cosθ2=π(12θ120)sinθ2cosθ2=π(12θ120)


Solution

Let the radius be r
BOC=θ
AOC=1800θ     [  AB is the diameter]
Area of minor segment cut by AC = Area of sector OADC - Area of
AOC
180360θ360πr2r2sin(90θ2).cos(90θ2)
=r2[(12θ360)]πsin(θ2).cos(θ2)   ...(1)   [sin(90θ)=cosθ2]
[cos(90θ)sinθ]
Area of sector BOC =θ360×πr2   ...(2)
r2[(12θ360)πsin(θ2.cosθ2)]=2×θ360×πr2
π2θπ360θπ180=sinθ2.cosθ2
π[12θπ2θπ360]=sinθ2.cosθ2
π[12θπ360]=sinθ2.cosθ2
π[12θπ120]=sinθ2.cosθ2

 

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