Triangles in n-gons and more...

Because we're considering a regular n-gon, we can inscribe it in a circle. We can denote the arc length between two neighboring vertices of the n-gon as $\theta$ degrees. Call the triangle $ABC$ . By power of the point, we get that $\measuredangle ABC=\frac { a\theta}{ 2 }$ where $a$ is the number of sides of the n-gon contained by arc $BC$ . Thus, we find that the angles are in the same ratio as the number of sides of the n-gon contained by each arc of two vertices of the triangle. The ratio given cannot be simplified, so the answer is just $2015$ . (This is just letting one arc contain two sides, another $1006$ and another $1007$ .)