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

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.

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.)