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Question

The inradius of pedal triangle of ΔABC is:

A
R2
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B
RsinAsinBsinC
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C
2RcosAcosBcosC
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D
4RsinA2sinB2sinC2
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Solution

The correct option is C 2RcosAcosBcosC
We know, sides of the pedal triangle are acosA,bcosB,ccosC.
Now, acosA=2RsinAcosA=Rsin2A
Area of pedal triangle is Δp=12DE.DF.sin(EDF)=12(Rsin2B)(Rsin2C)sin(1802A)=12R2sin2Asin2Bsin2C
Now, semi perimeter is Sp=12(Rsin2A+Rsin2B+Rsin2C)=R2(sin2A+sin2B+sin2C)=2RsinAsinBsinC
Hence inradius rp=ΔpSp=Rsin2Asin2Bsin2C4sinAsinBsinCrp=2RcosAcosBcosC

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