25. Two spheres are "inscribed" in a regular truncated triangular pyramid: one touches all its faces, the other touches all its edges.

Calculate the dihedral angle between the planes of the base of the pyramid and its lateral face.
25. Let $a$ denote the length of the side of the upper base, and $b$ - the length of the side of the lower base of the pyramid $(a<b)$. The sphere that touches the edges of the pyramid intersects the lateral face (an isosceles trapezoid) along a circle inscribed in this face. By the property of the sides of a circumscribed quadrilateral, the lateral sides of this trapezoid have a length of $\frac{a+b}{2}$, from which it is easy to find that the apothem of the truncated pyramid is $\sqrt{a b}$. The sphere inscribed in the pyramid touches the bases at their centers and the lateral faces at points on their apothems. By making a section through the center of the inscribed sphere and the apothem of the pyramid, we obtain a circle in the section that touches the bases and one of the lateral sides of the trapezoid - the section of the pyramid by a plane. In the plane of the section, draw a tangent to this circle, symmetric to the apothem relative to the height of the pyramid passing through the center of the sphere, and we get a circle inscribed in an isosceles trapezoid. The bases of this trapezoid are easily seen to be $\frac{a}{\sqrt{3}}$ and $\frac{b}{\sqrt{3}}$, so its height (the diameter of the inscribed circle) is $\sqrt{\frac{a b}{3}}$. Since the apothem of the pyramid, i.e., the height of the lateral face, is $\sqrt{a b}$, the sine of the dihedral angle between the lateral face and the base is $\sin \alpha=\frac{\sqrt{\overline{a b}}}{\sqrt{\overline{3} a b}}=\frac{1}{\sqrt{3}}$, from which $\alpha \approx 35^{\circ} 15^{\prime} 52^{\prime \prime}$.