2. Find all triples ( $a, b, c$ ) where $a, b, c$ are the lengths of the sides of triangle $ABC$ with angles $\alpha, \beta, \gamma$, such that the numbers $\cos \alpha, \cos \beta, \cos \gamma$ are the lengths of the sides of a triangle congruent to triangle $ABC$.
Solution. From the conditions of the problem, it follows that $\cos \alpha, \cos \beta, \cos \gamma$ are positive. Therefore, triangle $ABC$ is acute-angled. Let $a \leq b \leq c$. Then $\alpha \leq \beta \leq \gamma$, so $\cos \gamma \leq \cos \beta \leq \cos \alpha$. From the condition of the problem, it follows that $a=\cos \gamma$ and $c=\cos \alpha$. Therefore, $a: c=\cos \gamma: \cos \alpha$. From the sine theorem, we get $a: c=\sin \alpha: \sin \gamma$, so from this and the previous equality, we have

$$
\cos \gamma: \cos \alpha=\sin \alpha: \sin \gamma \text {, i.e., } \sin \alpha \cos \alpha=\sin \gamma \cos \gamma \text {. }
$$

Thus $\sin 2 \alpha=\sin 2 \gamma$ or $2 \alpha+2 \gamma=\pi$ (this is not possible, because then $\beta=\frac{\pi}{2}$, so $\cos \beta=0$). Therefore, $\alpha=\gamma$, i.e., $\alpha=\beta=\gamma=\frac{\pi}{3}$, so there is only one triplet with the desired property. This triplet is ( $\frac{1}{2}, \frac{1}{2}, \frac{1}{2}$ ).