2. Determine the angle $\alpha$ if

$$
\operatorname{ctg} \alpha=2+\sqrt{a}+\sqrt{b}+\sqrt{c}, \operatorname{ctg} 2 \alpha=2+\sqrt{a}
$$

where $a, b, c$ are natural numbers not divisible by 4, and $\sqrt{a}, \sqrt{b c}$ are irrational numbers.
Solution. From the given assumptions, we have

$$
2+\sqrt{a}=\operatorname{ctg} 2 \alpha=\frac{\operatorname{ctg}^{2} \alpha-1}{2 \operatorname{ctg} \alpha}=\frac{(2+\sqrt{a}+\sqrt{b}+\sqrt{c})^{2}-1}{2(2+\sqrt{a}+\sqrt{b}+\sqrt{c})}
$$

from which, after simplification, we get

$$
4 \sqrt{a}-2 \sqrt{b c}=b+c-a-5
$$

The right-hand side of the last relation is an integer, let's denote it by $d$. From $4 \sqrt{a}-2 \sqrt{b c}=d$ it follows that $4 d \sqrt{b c}=4 b c+d^{2}-16 a$, and since by assumption the number $\sqrt{b c}$ is irrational, the only possibility is $d=0$. Further, we get $4 a=b c$ and $a+5=b+c$. The numbers $b$ and $c$ are not divisible by 4, so from $4 a=b c$ it follows that $b=2 b^{\prime}$ and $c=2 c^{\prime}$, where $b^{\prime}, c^{\prime}$ are odd integers. Now we get $a=b^{\prime} c^{\prime}$ and $a+5=2\left(b^{\prime}+c^{\prime}\right)$, from which by eliminating $a$ we get $c^{\prime}\left(b^{\prime}-2\right)=2 b^{\prime}-5$. However, $b^{\prime}$ is odd, so $b^{\prime} \neq 2$ and from the last equality it follows that $c^{\prime}=2-\frac{1}{b^{\prime}-2}$. For $c^{\prime}$ to be an integer, it is necessary and sufficient that $b^{\prime}=1$ or $b^{\prime}=3$. It is easily verified that in these cases the corresponding triples are $a=3, b=6, c=2$, or $a=3, b=2, c=6$, which satisfy the conditions of the problem.

In both cases $\operatorname{ctg} 2 \alpha=2+\sqrt{3}$, from which we get $\alpha=\frac{\pi}{24}+k \pi, k \in \mathbb{Z}$.