Mathematical properties of two dimensional projection of three dimensional rotated object Please be gentle as I do not have any degree in maths.
By using a compass/straighedge method to construct Metatron's cube, a regular dodecahedron can be inferred from intersecting points. I'm looking for the ratio between the lengths of the edges (blue) of the dodecahedron and the radius of the initial circle (red) used for the construction.
What I actually want is to have on of the faces of the dodecahedron, to be a regular pentagon (purple) on the two dimensional plane on which it's being projected. If you take a horizontal line through the center of the dodecahedron and rotate the object over that line (green). 
1. How many degrees does it need to be rotated to make the irregular pentagon below, regular? 

2. Is it true that after this rotation, the circle is perfectly inscribed inside the pentagon?

EDIT: please understand that the purple pentagon only appears after a rotation of the resulting dodecahedron in 3D space 




*Can I then say that if $r = 1$ then $x = 1.45308505601$ by using the formula for calculating the apothem ($DB$) given the length of a side which is: $$y = \frac{s}{2tan\frac{180}{5}} $$


For $y=1$ that gives $1.45308505601$.
(ref: http://www.mathopenref.com/apothem.html)
 A: Your construction does not yield the projection of a regular dodecahedron onto a plane. Take for instance the upmost pentagon in your figure: its horizontal side is parallel to the plane, so its length (which can be calculated to be 
$4\sqrt3/5$) should be the actual length of every edge of the dodecahedron. But if so the other sides of that pentagon should be shorter than the horizontal one, which is not the case. In addition, the length of a pentagon circumscribed to a unit radius circle is $2\sqrt{5-2\sqrt5}$, which is different from $4\sqrt3/5$.
In any case the rotation angle $\theta$ you need (for a real dodecahedron) is just $\angle AOB$, where $O$ is the dodecahedron centre, $A$ is the centre of the blue pentagon and $B$ is the vertex at the center of your diagram. Using the coordinates of the vertices which can be found here, one can compute (warning: a previous value was not correct):
$$
\cos\theta={1\over\sqrt{15+6\sqrt5}},
\quad\hbox{that is}\quad 
\theta\approx 79.188°.
$$

EDIT.
To better explain my above remarks, you can see below a diagram made with GeoGebra. The blue edge of the dodecahedron in space is parallel to the projection plane, hence its projection should have the same length (0.34641 in picture below, where the external hexagon has unit side). The green edge, on the other hand, is not parallel to the projection plane and the length of its projection should be less than its length in space. But in your construction the length of the supposed projection is 0.35277, which is greater. It follows that the construction doesn't reproduce faithfully a dodecahedron projection.

