6.24. Three dihedral angles of a tetrahedron, not belonging to the same vertex, are equal to $90^{\circ}$, and all other dihedral angles are equal to each other. Find these angles.

## § 4. Equifacial Tetrahedron

Definition. A tetrahedron is called equifacial if all its faces are equal, i.e., its opposite edges are pairwise equal.
6.24. As follows from the solution of problem 6.22, the vertices of the given tetrahedron are the vertices $A, B, D$ and $D_{1}$ of the rectangular parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$. Let $\alpha$ be the required angle; $A B=a, A D=b$ and $D D_{1}=c$. Then $a=$ $=b \operatorname{tg} \alpha$ and $c=b \operatorname{tg} \alpha$. The cosine of the angle between the planes $B B_{1} D$ and $A B C_{\mathbf{1}}$ is

$$
a c / \sqrt{a^{2}+b^{2}} \sqrt{b^{2}+c^{2}}=\operatorname{tg}^{2} \alpha /\left(1+\operatorname{tg}^{2} \alpha\right)=\sin ^{2} \alpha
$$

(see problem 1.9, a). Therefore, $\cos \alpha=\sin ^{2} \alpha=1-\cos ^{2} \alpha$, i.e., $\cos \alpha=(-1 \pm \sqrt{5}) / 2$. Since $1+\sqrt{5}>2$, we finally obtain $\alpha=\arccos ((\sqrt{5}-1) / 2)$.