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Question

In a ΔABC,tanA:tanB:tanC=1:2:3 Hence sinA:sinB:sinC=k:m:n. Find m2k2n ?

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Solution

tanA=K,tanB=2K,tanC=3K
tanA=tanA6K=6K31=K2K=1
sinA=12,sinB=25,sinC=310sinA:sinB:sinC=12:25:310=5:22:3
m2k2n=853=0

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