Why do we divide by $ \sqrt{A^2 + B^2} $ to convert a function from the standard form to the normal form? To convert an equation in the standard form, $ Ax + By + C = 0 $ to the normal form $ \cos \omega \ x + \sin \omega \ y = p $, we first divide by $ \sqrt{A^2 + B^2} $. Why? 
For example, to convert the equation $ \sqrt 3 \ x + y - 8 = 0 $, we divide by $ \sqrt{A^2 + B^2} = 2 $ to get $ \frac{\sqrt 3}{2} \ x + \frac 12 \ y = 4 $ and then solve for $ \cos \omega = \frac{\sqrt 3}{2} $ and $ \sin \omega = \frac 12 $to find the value of $ \omega $. 
I understand that we can't possibly convert this directly, since there is no $ \omega $ for which $ \cos \omega = \sqrt 3 $--but why do we specifically divide by $ \sqrt{A^2 + B^2}?$ Why not any other number to bring the values of $ \cos \omega $ and $ \sin \omega $ between 0 and 1 and then solve it?

We can solve $(\cos\omega,\sin\omega)=(a,b)$ if and only if $(a,b)$ is a unit vector (i.e. the point $(a,b)$ is on the unit circle). The original vector $(A,B)$ is not unit, so we need to normalize it, that is, to find another vector of the length one that has the same direction. It is done by dividing the vector by its length
$$
(a,b)=\frac{(A,B)}{\|(A,B)\|}=\frac{(A,B)}{\sqrt{A^2+B^2}}.
$$
