1thr can be some shorter proof... but i jus went on doing without thinking and landed up with this result.... :)
in triangle abc,Δ2=a2b2c2/2(a2+b2+c2)then which type triangle is abc right angel, equilateral,isoscelesor scalene
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3 Answers
62a2+b2+c2=8R2
It is right angled.. but how i will have to think a bit more :)
basically bcos c=2R for a right angled traingle
and a2+b2=c2
But this is not a proof but a justification....
I will try to get a proof..
1a^2+b^2+c^2=8R^2 -----------(1)
from sine law,,, a=2RsinA, b=2RsinB, c= 2RsinC...
puting this in eqn 1,,,
sin^2A+ sin^2B+ sin^2C = 2
=>1- 2sin^2A+1- 2sin^2B+ 1- 2sin^2C = 3 - 2X2 =-1
=>cos2A+ cos2B + cos2C +1 =0
=>2cos(A+B)cos(A-B) + cos[2(Ï€-(A+B))] +1 =0
=>2cos(A+B)cos(A-B) + cos2(A+B) +1 =0
=>2cos(A+B)cos(A-B) + 2cos^2(A+B) -1 +1 =0
=> cos(A+B) [cos(A-B) + cos(A+B)] =0
=>cos(A+B)[2cosAcosB] =0
=> either cos(A+B)=0 OR cosA=0 or cosB=0
=> A+B =90° OR A=90° OR B=90°....
HENCE IT'S A RIGHT TRIANGLE !!! :)
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