Maximum number of acute triangles in a regular convex polygon triangulated into $n-2$ triangles by its diagonals. I have been reading about convex polygons, and I found the following:

We say that a simple polygon is convex if all its interior angles are less than $\pi$. If $P$, a regular convex polygon, is divided into $n-2$ triangles with diagonals, what is the maximum number of acute triangles one can have?

I don't understand what is meant by "$n-2$ triangles with diagonals".
Thank you.
Note: I have understood what you mean by "$ n-2 $ triangles with diagonals", but I have tried a lot to solve it using everything that has been written to me, and I have not succeeded, I do not know how to take the factor that the angles have to be sharp. I would appreciate it if you could help me solve it. Thank you.

To reword that sentence,

If $P$ is a convex polygon with $n$ sides then you can use several diagonals of $P$ to subdivide $P$ into $n-2$ triangles.

In fact we can be still more precise than that:

If $P$ is a convex polygon with $n$ sides then you can use $n-3$ diagonals of $P$ to subdivide $P$ into $n-2$ triangles.

For example, if $n=4$, and so $P$ is a quadrilateral, then you can use $1$ diagonal to subdivide $P$ into $2$ triangles.
Next, if $n=5$ and so $P$ is a pentagon, then you can use 2 diagonals of $P$ to subdivide $P$ into $3$ triangles.
And so on.
