in a triangle ABC ,if median and altitude drawn from the vertex A,then tan(b-c)is equal to
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Step-by-step explanation:
Let ABC be a triangle AH the height from A to BC and AM the median (M the midpoint between B and C) without loss of generality let's say that H is between B and M so we have ΔBAH and ΔMAH congruent as angles BHA and MHA are equal angle BAH and MAH are equal (per hypothesis of problem) and they share side AH. Therefore BM=HM(orHM=
2
1
BM)
So, Lets' see ΔAHC. AM is bisector therefore we have ratios
MC
MH
=
AC
AH
but MH=
2
1
MB=
2
1
MC
therefore
MC
MH
=
2
1
So AHC is a right angled triangle at H and AC=2AH. So, sinC=
2
1
and C=30
0
∠HAC=60
0
and ∠HAB=∠HAM=
2
1
∠HAC=30
0
So ∠BAC=90
0
So, ∠BAD=60
0
∠BAC=3×30
0
=90
0
.
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