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

## right angled but may not be isosceles equilateral right angled and isosceles isosceles but may not be right angled

Aug 5, 2018

Given

${\sin}^{2} A + {\sin}^{2} B + {\sin}^{2} C = 2$

$\implies 1 - {\sin}^{2} A + 1 - {\sin}^{2} B - {\sin}^{2} C = 0$

$\implies {\cos}^{2} A + {\cos}^{2} B - {\sin}^{2} C = 0$

$\implies 2 {\cos}^{2} A + 2 {\cos}^{2} B - 2 {\sin}^{2} C = 0$

$\implies 1 + \cos 2 A + 1 + \cos 2 B - 2 \left(1 - {\cos}^{2} C\right) = 0$

$\implies 1 + \cos 2 A + 1 + \cos 2 B - 2 + 2 {\cos}^{2} C = 0$

$\implies \cos 2 A + \cos 2 B + 2 {\cos}^{2} C = 0$

$\implies 2 \cos \left(A + B\right) \cos \left(A - B\right) + 2 {\cos}^{2} C = 0$

$\implies \cos \left(\pi - C\right) \cos \left(A - B\right) + {\cos}^{2} C = 0$

$\implies - \cos C \cos \left(A - B\right) + {\cos}^{2} C = 0$

$\implies \cos C \cos \left(A - B\right) - {\cos}^{2} C = 0$

$\implies \cos C \cos \left(A - B\right) - \cos C \cdot \cos \left(\pi - \left(A + B\right)\right) = 0$

$\implies \cos C \left[\cos \left(A - B\right) + \cos \left(A + B\right)\right] = 0$

$\implies \cos C \cdot 2 \cos A \cos B = 0$

So any of $A , B \mathmr{and} C$ must be ${90}^{\circ}$

If $A = {90}^{\circ}$ then ${\sin}^{2} A = 1$

And then $B + C = {90}^{\circ}$

So ${\sin}^{2} B + {\sin}^{2} C$

$= {\sin}^{2} \left(\frac{\pi}{2} - C\right) + {\sin}^{2} C$

$= {\cos}^{2} C + {\sin}^{2} C = 1$

Hence ${\sin}^{2} A + {\sin}^{2} B + {\sin}^{2} C = 2$ is satisfied for any right angled triangle.