39. The center of the lower base of the cube is connected by straight lines to the four vertices of the upper base of the cube. Calculate the angles between these lines.
39. Let's place the origin of the coordinate system at the center of the base of the cube, direct the axes $O x$ and $O y$ along the diagonals of the base, and take the length of half the diagonal of the base as the unit. Then the coordinates of the four vertices will be: $(1,0, \sqrt{2}),(0,1, \sqrt{2}),(-1,0, \sqrt{2})$, $(0,-1, \sqrt{2})$. Using the general formula for the cosine of the angle between vectors, we get for the vectors going to the adjacent vertices:

$$
\cos \varphi=\frac{1 \cdot 0+0 \cdot 1+\sqrt{2} \cdot \sqrt{2}}{\sqrt{3} \cdot \sqrt{3}}=\frac{2}{3}
$$

Thus, $\varphi \approx 48^{\circ} 11^{\prime} 23^{\prime \prime}$.

For vectors directed to opposite vertices,

$$
\cos \varphi^{\prime}=\frac{1 \cdot(-1)+0 \cdot 0+\sqrt{2} \cdot \sqrt{2}}{\sqrt{3} \cdot \sqrt{3}}=\frac{1}{3} \text { and } \varphi^{\prime} \approx 70^{\circ} 31^{\prime} 44^{\prime \prime}
$$