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Question

If perimeters of LMN and ABC are λ and μ, then the value of λμ is:

A
rR
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B
Rr
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C
rR
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D
rR
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Solution

The correct option is A rR
Here, MNL is pedal triangle, then
MN=acosA,NL=bcosB and ML=ccosC
and MLN=π2A,LMN=π2B, and MNL=π2C
λ=MN+NL+LM
=acosA+bcosB+ccosC
Using sine rule,for a,b,c we have
=2RsinAcosA+2RsinBcosB+2RsinCcosC
Using multiple angle formula 2sinθcosθ=sin2θ we get
=R(sin2A+sin2B+sin2C)
Consider, sin2A+sin2B+sin2C
=2sin(A+B)cos(AB)+2sinCcosC
=2sin(πC)cos(AB)+2sinCcosC
=2sinCcos(AB)+2sinCcosπ(A+B)
=2sinC[cos(AB)cos(A+B)]
On applying transformation formula above we get
=2sinC[2sinAsinB]
=4sinAsinBsinC
Substituting for sin2A+sin2B+sin2C=4sinAsinBsinC we get
λ=R(sin2A+sin2B+sin2C)
=4RsinAsinBsinC
Using sine rule, we have
=4Ra2R×b2R×c2R
=abc2R2
=4R2R2
=2R
and μ=2s=2r
λμ=2R2r=rR
986638_1090641_ans_f865c246b7b046fd8bf6761025399eaf.png

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