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

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

Grade:11

2 Answers

Nishant Vora IIT Patna
askIITians Faculty 2467 Points
7 years ago
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
mycroft holmes
272 Points
7 years ago
Let tan A = k, then tan B = 2k, and tan C = 3k.
 
In a triangle tan A+tan B +tan C = tan A tan B tan C and so we get k=1.
 
So tan A =1, tan B = 2, tan C = 3. so that 
 
So a2:b2:c2 = sin2A: sin2B:sin2C =
 
 \frac{1}{2}: \frac{2}{5}: \frac{3}{10} = 5:2:3

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