Example 5: Through the vertex $A$ of the 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 $\qquad$.
Explanation: Let the edge length of a regular tetrahedron be 1. Draw $AO \perp$ plane $BCD$ at point $O$, then $AO=\frac{\sqrt{6}}{3}$. With $O$ as the center and $\frac{\sqrt{6}}{3} \cot 75^{\circ}$ as the radius, draw a circle on plane $BCD$. It is easy to see that this circle is inside $\triangle BCD$, and the intersection line of the desired section with plane $BCD$ is a tangent to this circle.

When the tangent is parallel to one side of $\triangle BCD$, the corresponding section $\triangle AMN$ is an isosceles triangle, and there are 6 such sections.
When $CB_1 = DC_1$, and $B_1C_1$ is tangent to the circle (as shown in Figure 6), the corresponding section $\triangle AB_1C_1$ is an isosceles triangle, and there are 6 such sections.

Draw $BE$ tangent to the circle, intersecting $CD$ at point $E$. By $\triangle BCE \cong \triangle ACE$, we get $BE = AE$, and the corresponding section $\triangle ABE$ is also an isosceles triangle, and there are 6 such sections.
In summary, there are 18 sections that meet the conditions.