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Question

If in a ABC,a2+b2+c2=8R2, where R= circumradius,then the triangle is

A
equilateral
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B
isosceles
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C
right angled
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D
none of these
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Solution

The correct option is A right angled
Since, a2+b2+c2=8R2
Using sine rule, we have
(2RsinA)2+(2RsinB)2+(2RsinC)2=8R2
4R2[(sinA)2+(sinB)2+(sinC)2]=8R2
sin2A+sin2B+sin2C=2
1cos2A+sin2B+1cos2C=2
cos2A+sin2Bcos2C=22=0
cos2Asin2B+cos2C=0
cos(A+B)cos(AB)+cos2C=0
Since, A+B+C=πA+B=πC
cos(πC)cos(AB)+cos2C=0
cosCcos(AB)+cos2C=0
cosC[cos(AB)+cosC]=0
cosC=0,cos(AB)+cosC=0
C=π2 or cos(AB)+cos(π(A+B))=0
cos(AB)cos(A+B)=0
Using transformation angle formula, we have
2cosAcosB=0
cosA=0,cosB=0
Hence A=π2or B=π2, or C=π2

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