The base of the pyramid $ABCDE$ is the square $ABCD$, and its side faces are equilateral triangles. The plane $S$ passing through the edge $AB$ divides the volume of the pyramid in half. What angle does $S$ form with the base of the pyramid?
Let's use the notations from Figure 1! $P, Q$ are the intersection points of the plane $S$ and the edges $ED, EC$; $T_{P}, T_{Q}$ are the orthogonal projections of points $P, Q$ onto the base of the pyramid; $F_{1}, F_{2}, F_{3}$ are the midpoints of segments $AB, CD, PQ$ respectively.

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Figure 1

The pyramid is cut by planes passing through points $P$ and $Q$, perpendicular to the base and parallel to the edge $AD$. The cut-off part consists of a frustum with a triangular base and two congruent but mirror-symmetric pyramids with rectangular bases. We will calculate the volumes of the frustum and the pyramids, and then use these quantities to determine the desired angle.

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Figure 2

Consider the plane section $E F_{1} F_{2}$ of the pyramid (Figure 2). Choosing the edge length of the pyramid to be 1, we have $E F_{1}=E F_{2}=\frac{\sqrt{3}}{2}$, $E E_{T}=\frac{\sqrt{2}}{2}$, where $E_{T}$ is the orthogonal projection of point $E$ onto the base of the pyramid. The orthogonal projection of point $F_{3}$ onto the base of the pyramid lies on segment $F_{1} F_{2}$, denoted by $T$. Let $F_{3} T=m, T F_{2}=r$. Then, due to the regularity of the pyramid, the distances of points $T_{P}$ and $T_{Q}$ from edges $AD$ and $BC$ are also $r$. Therefore, the base area of the frustum formed by cutting the lower part of the pyramid is $\frac{1}{2} \cdot 1 \cdot m$, its height is $1-2r$, and thus its volume is: $V_{1}=\frac{(1-2r) \cdot m}{2}$. The volume of the two cut-off pieces reassembled into a pyramid is:

$$
V_{2}=\frac{1 \cdot 2r \cdot m}{3}=\frac{2r \cdot m}{3}
$$

The volume of the original pyramid is:

$$
V=\frac{1 \cdot 1 \cdot \frac{\sqrt{2}}{2}}{3}=\frac{\sqrt{2}}{6}
$$

According to the condition, the plane $S$ bisects the volume of the pyramid, so:

$$
V=\left(V_{1}+V_{2}\right) \cdot 2, \quad \text { or }: \quad \frac{\sqrt{2}}{6}=\left(\frac{(1-2r) m}{2}+\frac{2r m}{3}\right) \cdot 2
$$

The triangles $E E_{T} F_{2}$ and $F_{3} T F_{2}$ are similar, and thus

$$
\frac{m}{r}=\frac{E E_{T}}{E_{T} F_{2}}=\sqrt{2}, \quad \text { that is } \quad m=r \cdot \sqrt{2}
$$

Using this in (1), we get the quadratic equation $1=6r-4r^2$ for $r$. The roots of this equation are $r_{1}=\frac{3+\sqrt{5}}{4}, r_{2}=\frac{3-\sqrt{5}}{4}$, but since the value of $r$ cannot be greater than 1, only $r_{2}$ is a solution.

The angle $F_{3} F_{1} T$ is the same as the desired angle, and with the known value of $r$, we can easily calculate the tangent of this angle:

$$
\operatorname{tg} \alpha=\frac{m}{1-r}=\frac{r \cdot \sqrt{2}}{1-r}=\frac{\sqrt{2}(3-\sqrt{5})}{1+\sqrt{5}}=0.3339
$$

From this, the desired angle is: $\alpha=18^{\circ} 27^{\prime} 42^{\prime \prime}$.

This completes the solution of the problem.