How $\pi$, $3.1415…$ and $180^o$ are adaptive together?!

I planed following to compute the circle's circumference.

The circle's circumference finally can computable from: $$\lim_{\alpha\to0}{\frac{360^o}\alpha d} = 2\pi r$$ I don't want to follow above computation here but it's computable and I already have done it on a paper.

I just could not understand when and how $360^o$ changed itself to $\pi=3.1415...$?! i.e. it mutated itself from an angle to a distance!

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The Wikipedia article describing the relation between degrees and radians might be useful. – fuglede Jan 31 '14 at 17:19
$\pi$ isn't a distance, $r$ is. $\pi$ is just a ratio, which is precisely what you get when you divide one degree measurement by another. – MartianInvader Jan 31 '14 at 17:19
I think if you just consider $\frac{360^\circ}\alpha$ as the number of triangles, the question disappears. – Carsten S Jan 31 '14 at 17:31

Seems like the underlying misunderstanding here is the degrees/radians relationship that @fuglede hinted at in the comments. How it specifically comes into play is in the sine function. When you're working this problem, and you're using sine (or any trig function) you need to remember that the function that uses degrees as an input, let's call it $sin_d()$, is a different function from the function that uses radians, $sin()$. For example,

$\lim_{\alpha\to0}{\frac{sin(k\alpha)}\alpha} = k$

but

$\lim_{\alpha\to0}{\frac{sin_d(k\alpha)}\alpha} = \frac{k\pi}{180}$

That's why $360$ (degrees) is becoming $2\pi$ (radians).

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There are a few issues with what you have.

Why not try this: Take $n$ as the number of sides of your regular $n$-gon. Then your angle $\alpha = 360^{\circ}/n.$ Now you can express the length of $d$ in terms of $r$ and $\alpha$ through the Law of Sines:

$$\frac{\sin \left(90^{\circ} - \frac{\alpha}{2}\right)}{r} = \frac{\sin \alpha}{d}.$$

Then, you have $n$ of those lengths comprising the perimeter of your $n$-gon. Take $n \to \infty$.

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Sorry I could not understand how this change explains changing $360{^\circ}$ from left side of my limit to $2\pi$ at my limit's right side. – Yasser Zamani Jan 31 '14 at 17:47