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?
 A: 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 $A+B+C=A_1+B_1+C_1=\pi$, $$A=A_1+\frac\pi2=B_1+C_1=\pi-A_1.$$
A: 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 $\frac{3\pi}{4}$.
