ratio of "diameter" of a n-gon to perimeter

Question

So say I have a regular polygon with n sides, and I bisect the an angle E such and the line (EF). Assume line segment EF has length b, while the polygons side length is s. What is $b/(n*s)$, and as n approaches $\infty$ does $b/(n*s)$ approach $\pi$?

Answer 1

So what we need to do here, is to define our angles: $\theta$ and $ \phi $, where:
$ \theta = \frac{\pi \cdot (n-2)}{n} $ and $\phi =\pi-2\theta$.
Now, applying the law of sines:
$ \frac{\sin \theta}{\frac{1}{2}b} = \frac{\sin \phi}{s} $
Then, you're all set up to continue on for your calculations!