In triangle abs if tanA:tanB:tanC=1:2 :3 then a2 :b2 :c2=

Dear student,

tanA =Δ/ s (s-a)
tanB =Δ/ s (s-b)
tanC =Δ/ s (s-c)
Where s = a+b+c/2

henceΔ/ s (s-a) :Δ/ s (s-b) :Δ/ s (s-c) = 1:2:3
so,1/ (s-a) : 1/ (s-b) : 1/(s-c) = 1:2:3

s-a: s-b: s-c = 1:1/2 : 1/3

b+c-a : a+c-b : a+b -c = 1:1/2:1/3

now solve individually both pairs

Thanks