# Triangle with $\sin^2 A +\sin^2 B =\sin C$

I want to know the triangles which satisfies the following equation :

$\sin^2 A + \sin^2 B = \sin C$. Here $A$, $B$, $C$ are angles of a triangle.

If we let $a$, $b$, $c$ to be the lengths of edges corresponded to angles, then we multiply $(abc)^2$ in the above equation. So we can conclude that $C$ is not larger strictly than $\pi/2$

So problem is divided into two cases without a loss of generality

Case 1 : $A$, $B < \pi/2$. This case is easy. Simple computation shows that $C = \pi/2$

Case 2 : $A > \pi/2$ and $B < \pi/2$

In this case I got one solution : $B=C$ so that sin(C) is a root of the equation $4t^2 +4t-1=0$

So I have a question : Can we solve this case in general ?

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+1 for showing some thought on the problem – Ross Millikan Jan 23 '13 at 4:29

As $\sin C \gt 0$ you are only interested in the root $\sin C= \frac 1{\sqrt 2}-\frac 12 \approx 0.20711, C \approx 0.20862 \approx 11.95^\circ$ Solution thanks to Alpha
Thank you. By using alpha, I can draw the implicit plot of $\sin^2(B+C) + \sin^2(B) =$sin$(C)$. So in $BC$-plane the solution is a some curve from $(0,0)$ to $(0,\pi/2)$ when $A>\pi/2$. – HK Lee Jan 23 '13 at 4:38