How can I find the 3D rotation angle of an isoceles trapezoid?

Question

I have an isoceles trapezoid for which I know the lengths of all four sides and the inner angles:

                 :
      ---------------------
     /           :         \
    /            :          \
   /             :           \
  /              :            \
 /               :             \
---------------------------------
                 :

That trapezoid is rotated in 3D around the vertical central axis by an arbitrary number of degrees such that one side is closer to the observer.

How can I find that rotation angle?

It seems to me that the apparent lengths of the sides and the apparent angles will change. I'm struggling to turn that intuition into a means of calculating the rotation.

Answer 1

nvm, got it ... as it rotates, the apparent height will remain unchanged while the width will vary eventually reaching 0 when viewed edge on. Therefore, the ratio of height to width varies according to the cosine of the angle.

                 wt
      ---------------------
     /           :         \
    /            :          \
   /             : h         \
  /              :            \
 /               :             \
---------------------------------
                 wb

Given actual height $h$, actual bottom-width $wb$, and apparent bottom-width $wb'$, let $\frac{wb}{h}$ = 1 (normalized ratio viewed face-on), the viewing angle is calculated as: $$$\alpha = cos^{-1}(\frac{wb'}{h})$$

(You could use the top width instead of bottom width in the same manner.)