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 !!! :)
