1. Without using tables or a calculator, calculate the acute angle $\alpha$ if it is known that

$$
\operatorname{ctg} \alpha=2+\sqrt{2}+\sqrt{3}+\sqrt{6}
$$
Solution. We have

$$
\begin{aligned}
\operatorname{ctg} \alpha & =2+\sqrt{2}+\sqrt{3}+\sqrt{6}=\sqrt{2}(\sqrt{2}+1)+\sqrt{3}(\sqrt{2}+1) \\
& =(\sqrt{2}+1)(\sqrt{2}+\sqrt{3})=\frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}-1}=\frac{\frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{2}}{\frac{\sqrt{2}}{2}-\frac{1}{2}} \\
& =\frac{\sin 60^{\circ}+\sin 45^{\circ}}{\sin 45^{\circ}-\sin 30^{\circ}}=\frac{2 \sin \frac{15^{\circ}}{2} \cos \frac{15^{\circ}}{2}}{2 \sin \frac{15^{\circ}}{2} \cos \frac{75^{\circ}}{2}}=\operatorname{ctg} \frac{15^{\circ}}{2}
\end{aligned}
$$

Let's note that when simplifying, we use the fact that

$$
\sin \frac{105^{\circ}}{2}=\cos \left(90^{\circ}-\frac{105^{\circ}}{2}\right)=\cos \frac{75^{\circ}}{2}
$$

Since $\alpha$ is an acute angle and $\operatorname{ctg} \alpha=\operatorname{ctg} 7^{\circ} 30^{\prime}$, we get $\alpha=7^{\circ} 30^{\prime}$.