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?

All of the sines are positive, because angles in a triangle are between $0$ and $\pi$. Consequently, the angles $A'$, $B'$, and $C'$ have positive cosines and are therefore all acute.
Every triangle has at least two acute angles, so without loss of generality, assume that in $\Delta ABC$, angles $A$ and $B$ are acute. (If not, switch the label of the obtuse angle with the label $C$ and make the corresponding switch for $\Delta A'B'C'$.) For acute angles $\theta$ and $\phi$, $\sin \theta = \cos \phi$ only if $\theta+\phi=\frac{\pi}{2}$. This means that $A=\frac{\pi}{2}-A'$ and $B=\frac{\pi}{2}-B'$.
Angle $C$ cannot be acute, because if it were, $C'$ would be $\frac{\pi}{2}-C$, and $A'+B'+C'$ (which equals $\pi$) would equal $\frac{3\pi}{2}-(A+B+C)$, which equals $\frac{\pi}{2}$.
Therefore $C$ is obtuse, and $\sin C=\cos C'$ implies that $\color{blue}{C=\frac{\pi}{2}+C'}$. Also note that $C$ is the only one of the six angles that is obtuse, so it’s the largest angle of the six.
But we also know that $\color{blue}{C=}\pi-A-B=\pi-(\frac{\pi}{2}-A')-(\frac{\pi}{2}-B')=A'+B'=\color{blue}{\pi-C'}$.
If we add the two blue expressions, we find that $2C=\frac{3\pi}{2}$, and therefore $C$, the largest of the angles, is $\frac{3\pi}{4}$.
