44. Find the dihedral angle between the base and the lateral face of a regular truncated triangular pyramid, given that a sphere can be inscribed in it and, moreover, there exists a sphere that touches all its edges.
44. Extend the lateral faces until they intersect. In this case, we obtain two similar pyramids, the bases of which are the larger and smaller bases of the given truncated pyramid. Let $a$ be the side of the larger base of the truncated pyramid, $\alpha$ be the dihedral angle at this base: We can find: the height of the larger pyramid $-h=\frac{a \sqrt{3}}{6} \operatorname{tg} \alpha$, the radius of the inscribed sphere $r=\frac{a \sqrt{3}}{6} \operatorname{tg} \frac{\alpha}{2}$, the height of the smaller pyramid $-h_{1}=h-2 r=$ $=\frac{a \sqrt{3}}{6}\left(\operatorname{tg} \alpha-2 \operatorname{tg} \frac{\alpha}{2}\right)$, the side of the smaller base $-a_{1}=$ $=\frac{h_{1}}{h} a=a \frac{\operatorname{tg} \alpha-2 \operatorname{tg} \frac{\alpha}{2}}{\operatorname{tg} \alpha}$, the lateral edge of the larger pyramid $l=\frac{a \sqrt{3}}{6} \sqrt{\operatorname{tg}^{2} \alpha+4}$, the lateral edge of the smaller pyramid $-l_{1}=$ $=l \frac{h_{1}}{h}$; after this, we will use the condition for the existence of a sphere that touches all the edges of the truncated pyramid. This condition is equivalent to the existence of a circle inscribed in the lateral face, i.e., the equality must hold

$$
2\left(l-l_{1}\right)=a+a_{1}
$$

Expressing $l, l_{1}, a_{1}$ in terms of $a$ and $\alpha$, we obtain the equation

$$
\frac{\sqrt{3}}{3} \sqrt{\operatorname{tg}^{2} \alpha+4} \cdot \operatorname{tg} \frac{\alpha}{2}=\operatorname{tg} \alpha-\operatorname{tg} \frac{\alpha}{2}
$$

From this, we find $\operatorname{tg} \frac{\alpha}{2}=\sqrt{3}-\sqrt{2}$.

Answer: $2 \operatorname{arctg}(\sqrt{3}-\sqrt{2})$.