So, first a disclaimer. The oblique shock charts are based on inviscid flow. This works great away from the wall, but exactly at the corner (if you zoom in like you're saying), there is of course a boundary layer. That boundary layer messes with reality.
If we stick with inviscid flow, then the streamline at the wall is no different from the rest of the streamlines. The inviscid wall is what we call a slip wall (a viscous wall is a no-slip wall).
The wall is just like another streamline. It is a streamline that turns twenty degrees. The next one out must do the same thing, etc. If you accept that the other streamlines turn and form a shock, then the same thing applies to the wall.
In this case, the angle of the oblique shock is strictly determined by \theta and Mach number.
Post Script
If that answer isn't satisfactory (which it probably isn't), then to really get into it, you have to go through the derivation and understand all the math.
It is only conservation of mass, energy, and momentum -- with a perfect gas equation of state. But connecting the dots from those simple principles to the end state is non trivial.
You need to read a textbook on gas dynamics or an aerodynamics textbook. Either one should do a good job with the oblique shock derivation. Don't just jump to that chapter, start at the beginning (of the compressible flow part if an aerodynamics book) and go through 1D flow, normal shocks, and then oblique shocks.
Edit: Trying to explain the shock angle a little more...
Here is something that blew my mind when I learned it.....
When you consider an oblique shock, you decompose the velocity vectors into two components -- the component normal to the shock and the component tangential (parallel) to the shock. (Don't worry about the fact that we don't know what the shock angle is yet).
Here the incoming velocity is V1. Like usual, the shock angle is $\beta$. The component of V1 normal to the shock is $V_{n1}$ and the component of V1 tangent to the shock is $V_{t1}$.
Likewise, after the shock, the velocity is V2, with components $V_{n2}$ and $V_{t2}$.
Here is the mind blowing part...
$$V_{t2}=V_{t1}$$
The tangential velocity after the shock is the same as the tangential velocity before the shock!
Here is the next mind blowing part...
In the direction normal to the shock -- the flow behaves exactly as if it was going through a normal shock!
The normal Mach number before the shock $M_{n1}$ is greater than 1.0 (supersonic) and the normal Mach number after the shock $M_{n2}$ is subsonic. The pressure, density, and temperature jumps all behave exactly like a normal shock as if the tangential velocity was not there.
Here is another view that may help...
If you were observing in a coordinate system attached to the shock, with the flow moving by you left to right, this is what you would see.
Next, imagine you were moving vertically at $V_{t1}$ such that $V_{t1}$ appeared to be zero to you. You would see exactly the normal shock situation. This is what the flow normal to the shock experiences.
So, if your guess was right (about the shock being parallel to the ramp), then $V_{n2}$ would have to be zero. The flow can't penetrate the ramp.
This is what I meant by 'where would the flow go'?
$V_{n2}$ must be non-zero and $V_{t2}=V_{t1}$. Together, these two components of velocity must form a triangle at the angle $\theta$.
Remember when I said "Don't worry about the fact that we don't know what the shock angle is yet". This is where the math gets a bit tricky.
To solve this problem (figure out all the angles, etc), we end up with a system of multiple equations and multiple unknowns.
We know $M_1$ and $\theta$, but we don't know $\beta$. Since we don't know $\beta$, we can't decompose $V_1$ into $V_{n1}$ and $V_{t1}$
We know that $V_{n2}$ and $V_{t2}$ must make the angle $\theta$, but we don't know their magnitudes.
We know that $V_{n1}$ and $V_{t1}$ must be the legs of a right triangle with magnitude $V_1$.
Then we know the normal shock relations for how $V_{n2}$ must relate to $V_{n1}$.
All this is non-trivial to solve and doesn't really lend itself to a simple physically intuitive solution.
In the days before computers, solving this was a real pain. So, really smart people at NACA (the predecessor to NASA) did the calculations for us and produced the famous theta-beta-Mach chart. They also produced numeric tabulated results. These were printed in volumes like NACA 1135. Paper copies of these were distributed widely -- to libraries at universities and research institutions as well as companies working on aerospace problems.
There are plenty of online copies available, but they're typically scans made from an old microfiche. They'll do, but they're grainy and low resolution. If you can lay your hands on an original paper copy (say on eBay), you'll find that the charts were quite beautiful.
