Which force to resolve? 
In the above image the particle is effectively moving 'out of the page' in circular motion. We want to solve for $\alpha$, the slope is smooth, and the particle is moving neither up nor down the slope.
Which is correct? $R\cos \alpha = mg$ or $R = mg \cos \alpha$?
If we say $R\cos \alpha = mg$ and $R\sin\alpha = \frac{mv^2}{r}$ then we can solve for $\alpha$ by dividing. This gives the correct answer. However, I am having trouble understanding the first of these two equations. The problem is that, in a statics problem involving a mass sitting on a ramp (the same diagram without the $\frac{mv^2}{r}$) we would use the equation $R=mg\cos\alpha$, and I cannot see why there should be a difference.
 A: I'll answer since you've given it the old college try.
Start by defining a coordinate system in which x is towards the center of the circle and y is upwards. The components of the normal force are
$$R_x=R\sin\alpha,\qquad R_y=R\cos\alpha$$
Then Newton's second law gives you
$$\sum F_x=R\sin\alpha=ma_c=m\frac{v^2}{r}$$
$$\sum F_y=R\cos\alpha-mg=0$$
The second line tells you $R\cos\alpha=mg$ (your first equation), and dividing the first line by this gives you
$$\frac{R\sin\alpha}{R\cos\alpha}=\frac{mv^2}{rmg},$$
which is the correct answer.
Specifically: You know what the acceleration is here, it's $v^2/r$. That acceleration is due to the inwards component of the reaction force $R$. Newton's second law tells us the sum of all the forces is the (mass times the) acceleration. The rest is just math.
A: TL;DR: $R \neq mg\cos(\alpha)$, because there is a component of $\frac{mv^2}{r}$ in the same direction.
Relative to horizontal and vertical

The horizontal force must be equal to resultant force towards the center of the circle (for circular motion).
$$
R \sin(\alpha) = \frac{mv^2}{r} \\ 
$$
The particle is in equilibrium in the vertical direction, so
$$
R \cos(\alpha) = mg
$$
Then, like you said, you can divide to get the correct answer
$$
\frac{R\sin(\alpha)}{R\cos(\alpha)} = \frac{v^2}{gr}, \quad \tan(\alpha) = \frac{v^2}{gr} \quad \therefore \quad \alpha= \arctan(\frac{v^2}{gr})
$$
Relative to the slope

The horizontal force must be equal to the horizontal component of the resultant force towards the center of the circle.
$$
mg\sin(\alpha) = \ F_{x, net} =  \frac{mv^2}{r}\cos(\alpha) 
$$
Similarly for the vertical, the vertical forces are equal to the vertical component of the resultant force
$$
R - mg\cos(\alpha) =  \frac{mv^2}{r}\sin(\alpha)
$$
You can rearrange the horizontal component easily into the correct answer. 
$$
g\frac{\sin(\alpha)}{\cos(\alpha)} = g\tan(\alpha) = \frac{v^2}{r} \quad \therefore \quad \alpha= \arctan(\frac{v^2}{gr})
$$
So both methods of resolving are equivalent. 
A: I would normally solve this problem with the following diagram:

The reaction force is normal to the surface, and it's bigger than just the component of gravity because the force provided to the object adds another component. This is the dreaded "centrifugal force" that doesn't exist - except in a rotating frame of reference, where it is a very real "fictitious" force. Once you add that to the diagram, you see that there is a downward force, and a lateral force, that need to be countered by the net reaction force. 
You can't simply estimate the reaction force from the gravitational force alone - and that's why you were running into trouble.
See this link for some additional analysis of "banked turn" problems.
