Given right angled parallelepiped $ABCDA1B1C1D1$, with bases $ABCD$ and $A1B1C1D1$, which are squares with side $1$. if $\angle (B1C;D1A) = 60^\circ$ find the length of the surrounding edge (I'm not sure if this is the right term in English, those edges are $AA1,BB1,CC1,DD1$). Here is drawing:
I'm totally lost on this one.
You can draw $DA1$,which is parallel to $B1C$. This does not define which angle is $60^\circ$. If $E$ is the intersection of the new line and $AD1$, it could be $AEA1$ or $AED$ Either way you have the end face composed of two equilateral triangles and two $120^\circ-30^\circ-30^\circ$ triangles.