Example 2. Given that the radius of the circumscribed circle of a regular dodecagon is $a$, find the sum of all the diagonals (i.e., all the lines connecting non-adjacent vertices) of the regular dodecagon.
Let the sum of distances from one vertex to all other vertices be $\mathrm{s}_{1}$. According to the conclusion from the previous problem:
$$
\mathrm{s}_{1}=2 \mathrm{a} \cdot \operatorname{ctg} \frac{\pi}{2} \times \frac{\pi}{12}=2 \mathrm{a} \cdot \operatorname{ctg} \frac{\pi}{24}.
$$
$\mathrm{s}_{1}$ includes two edges connecting adjacent vertices, i.e., two adjacent sides of the regular dodecagon. Therefore, the sum of the distances from one vertex to all other non-adjacent vertices, $\mathrm{s}_{2}$, is:
$$
\mathrm{s}_{2}=2 \mathrm{a} \cdot \operatorname{ctg} \frac{\pi}{24}-2 \mathrm{a} \cdot \sin \frac{\pi}{12} \times 2.
$$

Here, $2 \mathrm{a} \cdot \sin \frac{\pi}{12}$ is the side length of the regular dodecagon.
Thus, the sum of the distances between all non-adjacent vertices in the regular dodecagon, $\mathrm{s}$, is:
$$
\mathrm{s}=\frac{12}{2} \cdot \mathrm{s}_{2}=12 \mathrm{a}\left(\operatorname{ctg} \frac{\pi}{24}-2 \sin \frac{\pi}{12}\right).
$$

Using relevant trigonometric formulas, the calculation yields:
$$
\mathrm{s}=6(4+3 \sqrt{2}+2 \sqrt{3}+\sqrt{6}) \mathbf{a}.
$$
(Author's affiliation: Xie Fami - Dangyang No.1 High School, Hubei Province; Xiong Sixing - Dangyang Textile Factory Technical Office, Hubei Province)