If in triangle ABC sin2A+sin2B+sin2C=2 then the triangle is options are below?

  1. right angled but may not be isosceles
  2. equilateral
  3. right angled and isosceles
  4. isosceles but may not be right angled

1 Answer
Aug 5, 2018

Given

sin2A+sin2B+sin2C=2

1sin2A+1sin2Bsin2C=0

cos2A+cos2Bsin2C=0

2cos2A+2cos2B2sin2C=0

1+cos2A+1+cos2B2(1cos2C)=0

1+cos2A+1+cos2B2+2cos2C=0

cos2A+cos2B+2cos2C=0

2cos(A+B)cos(AB)+2cos2C=0

cos(πC)cos(AB)+cos2C=0

cosCcos(AB)+cos2C=0

cosCcos(AB)cos2C=0

cosCcos(AB)cosCcos(π(A+B))=0

cosC[cos(AB)+cos(A+B)]=0

cosC2cosAcosB=0

So any of A,BandC must be 90

If A=90 then sin2A=1

And then B+C=90

So sin2B+sin2C

=sin2(π2C)+sin2C

=cos2C+sin2C=1

Hence sin2A+sin2B+sin2C=2 is satisfied for any right angled triangle.