θ = (length of arc)/(angle subtended by it). How? Thats it! Thats what I want to know. 
If θ is the angle subtended by an arc of length L at its center with radius R.
We know, θ = L/R.
How did we get this? 
Please don't say we got it from generalizing the idea of perimeter of circle.
 A: The length of the arc is proportional to the angle by proposition 6.33 in Euclid's Elements. We can choose units to make the constant of proportionality equal to $1$ for a circle of radius $1$. Those units are called radians.
[I give the Euclid reference since it uses the notion of angle that might be familiar to you. Nowadays we find it more economical to define the trignometric functions without any appeal to geometry (for eg. via the complex exponential), then define angles by inverting one of these functions, and then derive your result as an exercise in calculus.
Here is a simple hand-waving argument along those lines. Let OA and OB be two radii of a circle of radius $r$. Then by simple trignometry the length of the line segment AB is $2r\sin(\theta/2)$. For small enough $\theta$, $\sin(\theta/2) \approx \theta/2$ so the length of AB is approximately $r\theta$. In this case the length of AB is also close to the length of the arc AB hence that is also approximately $r\theta$. We can make the approximation as good as we like by making $\theta$ small enough. We can handle large arcs by approximating it by a large number of small segments of this sort.]
A: This is because using radian measure (as opposed to using degrees) in working with things related to circles is mathematically very convenient. In more advanced mathematical work you rarely see people use degrees since the formulae that have the angles in degrees require having to keep around inconvenient factors of $\frac{\pi}{180}$ or its reciprocal.
A: What is your definition of an angle? The formula you quoted follows directly from the definition of an angle as the length of the arc on the unit circle, combined with the intercept theorems (for “scaling the circle to a unit circle,” if you want).
Using angles defined this way is, as J.M. said, much more convenient than degrees. E.g., it connects triangles with trigonometric functions with nice derivatives – from $\sin' = \cos$, $\cos' = -\sin$, $\sin(0) = 0$, and $\cos(0) = 1$, you directly find that $\cos(\pi/2)=0$, i.e., these formulas only hold if the arguments to $\sin$ and $\cos$, traditionally angles, are given in radians.
