If in triangle ABC #sin^(2)A+sin^(2)B+sin^(2)C=2# then the triangle is options are below?

Question

right angled but may not be isosceles

equilateral

right angled and isosceles

isosceles but may not be right angled

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Answer 1 · 2018-08-05T03:33:23

Given

#sin^2A+sin^2B+sin^2C=2#

#=>1-sin^2A+1-sin^2B-sin^2C=0#

#=>cos^2A+cos^2B-sin^2C=0#

#=>2cos^2A+2cos^2B-2sin^2C=0#

#=>1+cos2A+1+cos2B-2(1-cos^2C)=0#

#=>1+cos2A+1+cos2B-2+2cos^2C=0#

#=>cos2A+cos2B+2cos^2C=0#

#=>2cos(A+B)cos(A-B)+2cos^2C=0#

#=>cos(pi-C)cos(A-B)+cos^2C=0#

#=>-cosCcos(A-B)+cos^2C=0#

#=>cosCcos(A-B)-cos^2C=0#

#=>cosCcos(A-B)-cosC*cos(pi-(A+B))=0#

#=>cosC[cos(A-B)+cos(A+B)]=0#

#=>cosC*2cosAcosB=0#

So any of #A, Band C# must be #90^@#

If #A=90^@# then #sin^2A=1#

And then #B+C=90^@#

So #sin^2B+sin^2C#

#=sin^2(pi/2-C)+sin^2C#

#=cos^2C+sin^2C=1#

Hence #sin^2A+sin^2B+sin^2C=2# is satisfied for any right angled triangle.