So why is normal force at inclined plane is $mg \cos a$ and friction is $mg \sin a$? I mean, why not vice-versa or why not some other ratio? Where does it come from? I can see that the sum squares of friction and normal force under the root should be equal to $mg$ and the angle at which the plane is inclined has something to do with it.
Does it come from some other part of physics or was it just deduced experimentally? I mean, we could slide a sample block from an inclined plane to see what acceleration it has each time we change the angle.
UPD:
Thank you all guys! What I was ultimately asking just seems to be a matter of philosophy. These all formulas are beautiful and seem to be correct. But they are abstractions. And the confidence that this is true ultimately comes from experiments. Since I am very new to physics, I just wanted to know if my assumptions were true.
Best Answer
In vector notation there is only one equation and no ambiguity. The block is in equilibrium so the net force acting on it must be zero. There are three forces acting on the block - its weight $\vec W$, the normal force $\vec N$ and friction $\vec F$. So we have
$\vec W + \vec N + \vec F = 0$
Since $\vec N$ and $\vec F$ are orthogonal to one another (at right angles) it is convenient to resolve the three vectors into components along an $x$ axis that is parallel to $\vec F$ and a $y$ axis that is parallel to $\vec N$. In component form we have
$\vec F = (F, 0) \\ \vec N = (0, N) \\ \vec W = (-W \sin (a), -W \cos (a))$
The last line comes from the fact that $\vec W$ is at an angle $a$ to the negative $y$ axis - it is simply the rule for how you turn a vector into components. Adding the three vectors and equating $x$ and $y$ components to zero gives two equations:
$F - W \sin (a) =0 \\ \Rightarrow F=W\sin (a)=mg\sin (a) \\ N - W \cos (a) =0 \\ \Rightarrow N=W\cos (a)=mg\cos (a)$
There is nothing special about this choice of $x$ and $y$ axes. We could instead choose axes that are horizontal and vertical. Relative to horizontal and vertical axes the component forms of the vectors are:
$\vec F = (F\cos (a), F\sin (a)) \\ \vec N = (-N\sin (a), N\cos (a)) \\ \vec W = (0, -W)$
If you add the three vectors and equate each component to zero you get two different equations. But if you then solve to find $F$ and $N$ in terms of $W$ you get exactly the same result as before - there are just a couple of extra steps in the algebra.