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.