8. Through the vertex $A$ of a regular tetrahedron $ABCD$, make a section in the shape of an isosceles triangle, and let the angle between the section and the base $BCD$ be $75^{\circ}$. The number of such sections that can be made is $\qquad$.
Answer
As shown in the figure, $A O \perp$ plane $B C D$, with $O$ as the center and $\frac{A O}{2+\sqrt{3}}$ as the radius, a circle is drawn. According to the problem, the intersection line of the section with the base $B C D$ is tangent to circle $O$. Draw a tangent from point $B$ to circle $O$, intersecting $C D$ at $E$, with the tangent point being $F$, and connect $A E$. From $\triangle A C E \cong \triangle B C E \Rightarrow A E=B E$ $\Rightarrow$ section $A B E$ meets the requirements, and such sections can be made 6; taking $B M=D N$ and making $M N$ tangent to circle $O$ $\Rightarrow$ section $A M N$ also meets the requirements, and such sections can also be made 6; obviously, the triangular sections formed by the tangents to circle $O$ parallel to the three sides of $\triangle B C D$ and point $A$ also have 6.
Therefore, there are a total of 18 sections that meet the requirements.