A line cuts off a triangle $A P Q$ from a regular $n$-sided polygon with side 1 such that $A P+A Q=1$ ( $A-$ is a vertex of the $n$-sided polygon).

Find the sum of the angles under which the segment $P Q$ is seen from all vertices of the $n$-sided polygon, except $A$.
Instead of changing the position of the vertex from which we look at the segment $Q P$, we can fix the vertex and change the position of the side (by rotating the $n$-gon). Thus, the sum we are interested in is equal to the sum of the angles under which all sides of a regular $n$-gon inscribed in the given one, one of whose sides is $-P Q$, are seen from point A. And this sum is equal to $\angle P A Q=\pi(n-2) / n$.

## Answer

$\pi(n-2) / n$