Question

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(12−θ120)sinθ2cosθ2=π(12−θ120)

sinθ2cosθ2=π(12−θ120)sinθ2cosθ2=π(12−θ120)

Solution

Let the radius be r

∠BOC=θ

∴∠AOC=1800−θ [∵ AB is the diameter]

Area of minor segment cut by AC = Area of sector OADC - Area of

∠ AOC

180360−θ360πr2−r2sin(90−θ2).cos(90−θ2)

=r2[(12−θ360)]π−sin(θ2).cos(θ2) ...(1) [∵sin(90−θ)=cosθ−2]

[cos(90−θ)sinθ]

Area of sector BOC =θ360×πr2 ...(2)

⇒r2[(12−θ360)πsin(θ2.cosθ2)]=2×θ360×πr2

⇒π2−θπ360−θπ180=sinθ2.cosθ2

⇒π[12−θπ−2θπ360]=sinθ2.cosθ2

⇒π[12−θπ360]=sinθ2.cosθ2

⇒π[12−θπ120]=sinθ2.cosθ2

∠BOC=θ

∴∠AOC=1800−θ [∵ AB is the diameter]

Area of minor segment cut by AC = Area of sector OADC - Area of

∠ AOC

180360−θ360πr2−r2sin(90−θ2).cos(90−θ2)

=r2[(12−θ360)]π−sin(θ2).cos(θ2) ...(1) [∵sin(90−θ)=cosθ−2]

[cos(90−θ)sinθ]

Area of sector BOC =θ360×πr2 ...(2)

⇒r2[(12−θ360)πsin(θ2.cosθ2)]=2×θ360×πr2

⇒π2−θπ360−θπ180=sinθ2.cosθ2

⇒π[12−θπ−2θπ360]=sinθ2.cosθ2

⇒π[12−θπ360]=sinθ2.cosθ2

⇒π[12−θπ120]=sinθ2.cosθ2

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