Question

If $\Delta$ ABC is a right-angled triangle prove that ${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=2$

Answer 1

Since $\Delta$ ABC is a right-angled triangle.

Let $C={90}^o$.........(1)

Then $B={90}^o-A$.......(2).

Now,

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

$={\sin }^{2}A+{\sin }^2({90}^o-A)+{\sin }^2{90}^o$

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

$=1+1$

$=2$