Agakhanov N.K.

The cosines of the angles of one triangle are respectively equal to the sines of the angles of another triangle.

Find the largest of the six angles of these triangles.
From the condition, it follows that the angles $\alpha_{1}, \alpha_{2}, \alpha_{3}$ of the first triangle are acute $\left(\cos \alpha_{i}=\sin \beta_{i}>0\right.$, where $\beta_{1}, \beta_{2}, \beta_{3}$ are the angles of the second triangle). Therefore, $\beta_{i}=90^{\circ} \pm \alpha_{i}$,

$i=1,2$, 3. From the equality $180^{\circ}=\beta_{1}+\beta_{2}+\beta_{3}=270^{\circ}+\left( \pm \alpha_{1} \pm \alpha_{2} \pm \alpha_{3}\right)$, where $\alpha_{1}+\alpha_{2}+\alpha_{3}=180^{\circ}$, it follows that there are both + and - signs in the parentheses.

Moreover, the second triangle cannot have two obtuse angles, so there is one + sign and two - signs in the parentheses.

Thus, up to permutation, $180^{\circ}=270^{\circ}+\left(\alpha_{1}-\alpha_{2}-\alpha_{3}\right)$, from which $\alpha_{1}=45^{\circ}$, that is, $\beta_{1}=135^{\circ}$. This is the only obtuse angle of the second triangle, and the first triangle is acute-angled; therefore, this angle is the largest.

## Answer

$135^{\circ}$.