Question

In triangle ABC, if the median and altitude drawn from vertex A, trisect angle A, then tan (B - C) is equal to.A B C are angles​

Answer 1

Given :   ΔABC, if the median and altitude drawn from vertex A, trisect angle A,

To Find :  tan | B - C |

Solution:

Let say AD ⊥ BC    &  AM  is median

=> BM  = CM   = BC/2

Let say  D lies between B & M  

∠BAD = ∠DAM = ∠MAC = α    and ∠A = 3α

in Δ ABD &  Δ AMD

∠BAD = ∠DAM  = α

AD = AD  ( common)

∠ADB =  ∠ADM = 90°

=> BD = DM

BD = DM = BM/2

=>  DM = CM/2  ( ∵ BM = CM)

in ΔADC

∠DAM = ∠MAC = α  hence AM is angle bisector

=> AD/AC  = DM/CM

=> AD/AC = 1/2

ADC  is right angled triangle

Hence CosC = AD/AC = 1/2

=> ∠C  = 30°

∠DAC = 60°

∠DAC =  ∠DAM + ∠MAC = 2α

=> α =30°

∠A = 3α = 90°

=> ∠B = 60°

tan | B - C |  = Tan |  60° - 30° |  = Tan 30°  = 1/√3

if D would have been between M & C

then ∠B = 30°  and ∠C = 60°

then tan | B - C |  = Tan |  - 30° |  = Tan 30°  = 1/√3

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