Hexagonal Albert-prism

Assuming unit sides for all the faces, $\overline{AB}=\frac{1}{2}$ . From symmetry, AC must be perpendicular to AD. Since $\angle BAD=60^\circ$ the foreshortened $\angle CAB=30^\circ$ . From the right $\triangle ABC$ we get the foreshortened $\overline{AC}=\frac{.5}{cos(30^\circ)}=\frac{1}{\sqrt{3}}$ . If we look at the quadrilateral face (which has to be vertical based on symmetry) head on, this is exactly the actual length of $\overline{EC}$ . The side of the quadrilateral, $\overline{AC}$ must in reality be 1. So from the right $\triangle AEC$ we get the $\angle EAC=arcsin(\frac{1}{\sqrt{3}})=35.26439^\circ$ . Doubling this will give us $70.529^\circ$ .