Beautiful little geometry problem about sines Given triangles ABC and $A_1B_1C_1$ such that $\sin A = \cos A_1, \sin B = \cos B_1, \sin C = \cos C_1$. What are the possible values for the biggest of these 6 angles?
I tried some stuff like sine theorem but can't derive from it? How do we do this one? Assume the closest to the smallest angle?

Short answer: $\frac{3\pi}{4}$.
Detailed answer: $A_1,B_1,C_1<\frac\pi2$. The equality
$$\sin A=\cos A_1$$
gives 
$$A=\frac\pi2-A_1\text{ or }A=\frac\pi2+A_1,$$
and similarly for $B$ and $C$. Working out the cases, the only possibility is
$$A=\frac\pi2+A_1,B=\frac\pi2-B_1,C=\frac\pi2-C_1.$$
Since $A+B+C=A_1+B_1+C_1=\pi$, $$A=A_1+\frac\pi2=B_1+C_1=\pi-A_1.$$
