Solution below.

Reveal solution by highlighting:

I had to fiddle a bit with this, to get the right approach.

If one of the sides of a triangle inside a regular polygon is a diameter of the polygon, then the triangle will have a right angle. The polygon has n vertices. There are n/2 distinct diameters. Every time I choose a diameter, I use up 2 vertices, so there’s only n-2 left. In all, I can create triangles in n/2*(n-2) different ways.

We know n/2*(n-2) = 14620. wolframalpha tells me, n = 172 is the only reasonable (i.e. positive) solution.