At what angle is a cube's width equal to its height, given an axonometric projection?

Question

Is it possible for the width of a 2D cube drawn using an axonometric projection (viewed at 45° around the vertical axis) to be equal to its "height" (actually the distance from the lowest to the highest vertical point)? How would I go about calculating the projection angle required to get this effect?

A) The 3D projection angles that produce the 2D cube

B) The resulting 2D cube whose flat height on the page needs to match to its flat width.

Answer 1

Assume a unit cube and a side view in the diagram, with the diagonal width between two sides $\sqrt2$. Then, the projection angle is

$$x =90 - 2\theta = 90-2\cot^{-1}\sqrt2 = \cot^{-1}(2\sqrt2)\approx 19.5^\circ$$