Why are triangles drawn like so when working with gravity on an inclined plane? This is my first year as a physics student, and I've never learned about vectors past a basic level, so this is confusing me. When we have gravity on an inclined plane, we separate it into two components, which I understand. 
However, consider the image below, and there's a box at point A. When separating the gravity components, you draw a triangle AGC (sorry, the G and D are on top of each other and difficult to distinguish). AG becomes the force of gravity in the y-direction, and GC becomes the force of gravity in the x-direction. Then you do the trig functions from there. However, when I tried this myself, I drew triangle ADF instead and tried the trig functions from there. It didn't work. I'm having trouble understanding why you can't compute the trig functions from AGF. The only partial solution I came up with was that force of gravity in the y-direction can't be the hypotenuse, as the force of gravity in the y-direction is always less than the force of gravity. But I think I'm missing something more.

 A: You can decompose the forces along AD and DF rather than AG and GC, but they won't be the relevant forces you're looking for. What makes this confusing is we often work entirely in magnitudes, whereas fundamentally we are manipulating vectors, and we only get away with this with good choices of decompositions.
Suppose the downward force of gravity is $\vec{f}$ with magnitude $f$ and direction along AC. The "right" method will say there is a normal force $\vec{f}_{\!\perp}$ in the direction of AG and a parallel force $\vec{f}_{\!\Vert}$ in the direction of GC, with $\vec{f} = \vec{f}_{\!\perp} + \vec{f}_{\!\Vert}$. The formulas for the magnitudes are
$$ f_{\!\perp} = f \cos\beta, \qquad f_{\!\Vert} = f \sin\beta. $$
Moreover, because AG and GC are orthogonal, we know $\vec{f}_{\!\Vert}$ cannot have any effect on pushing into the inclined plane; all such effects are captured by $\vec{f}_{\!\perp}$.
Now consider the "wrong" decomposition of $\vec{f} = \vec{f}_1 + \vec{f}_2$, with $\vec{f}_1$ along AD and $\vec{f}_2$ along DF. We can get these magnitudes too:
$$ f_1 = f \sec\beta, \qquad f_2 = f \tan\beta. $$
The problem is, $\vec{f}_1$ doesn't fully capture the normal force, because $\vec{f}_2$ contributes to this as well.
For an extreme example, imagine a mass sitting on horizontal ground with weight $\vec{f}$ directed downward. We could write $\vec{f} = \vec{f}_1 + \vec{f}_2$ with both $\vec{f}_1$ and $\vec{f}_2$ also pointing downward. We cannot just look at $\vec{f}_1$ and neglect $\vec{f}_2$ when considering the weight of the mass on the ground.
Another way of looking at things is that we are silently taking dot products. The real, unambiguous definition of the normal force of the mass on the block is the dot product of its weight vector with the unit normal vector to the surface, $\vec{f}_{\!\perp} = (\vec{f} \cdot \hat{n}) \hat{n}$ (give or take a sign). When computing dot products, you can only ignore components of $\vec{f}$ that are orthogonal to $\hat{n}$; if you do a decomposition where neither component is orthogonal, you have to include both terms. In equations, this is the difference between
$$ \vec{f} \cdot \hat{n} = \vec{f}_{\!\perp} \cdot \hat{n} + \vec{f}_{\!\Vert} \cdot \hat{n} = (f_{\!\perp}) (1) \cos0^\circ + (f_{\!\Vert}) (1) \cos90^\circ = f_{\!\perp}. $$
and
$$ \vec{f} \cdot \hat{n} = \vec{f}_1 \cdot \hat{n} + \vec{f}_2 \cdot \hat{n} = (f_1) (1) \cos0^\circ + (f_2) (1) \cos69.91^\circ. $$
A: Keep your goal in mind: to set up balances of forces in the $x$ and $y$ direction, which will normally be used later in the treatment of the problem to determine equations of motion. You haven't defined these axis yet.
You can choose them (your frame of reference) as you like, as long as they're perpendicular to each other. In this case the easiest choice is to choose AB as the $x$ axis, pointing in the B direction. Choose AG (which is perpendicular to AB) as the $y$ axis, pointing in the A direction. The origin of your $x,y$ coordinate system is A.
Call the angle at B, $\beta$. In this particular problem $\beta$ is the only angle you will need.
Assuming FB is the true horizontal, then the weight $mg$ of the box acts downwards along AC, perpendicular to FB.
The projection of $mg$ on AG is the $y$-component of $mg$ and is given by $-mg\cos\beta$ (minus sign in accordance with the sense of the $y$ axis).
The projection of $mg$ on AB is the $x$-component of $mg$ and is given by $mg\sin\beta$ (plus sign in accordance with the sense of the $x$ axis).
A: The problem can be worked in a x-horizontal y-vertical coordinate system but it is rather harder that way. The fundamental problem is the nature of the normal force: it is expressed as a constraint rather than some consistent rule like "$mg$ downward" for weight.
What the normal force does is resist attempts to push things into the same volume of space, so its direction is outward perpendicular to the surface and its magnitude is equal to whatever forces are trying to push the object together.
So you need to find "whatever forces are trying to push the objects together" to figure out what the normal force acts to cancel. In this case that means a portion of weight and nothing else, so you want to decompose the weight into the part trying to push the block into the ramp and whatever is left over (which is the "along the ramp" part).
