$24 \cdot 4$ A right circular cylinder with a base radius of 1 is cut by a plane, resulting in an elliptical cross-section. The eccentricity of this ellipse is $\frac{\sqrt{5}}{3}$. The acute angle formed between this cross-section and the base of the cylinder is
(A) $\frac{\pi}{4}$.
(B) $\frac{\pi}{3}$.
(C) $\arccos \frac{2}{3}$.
(D) $\arccos \frac{1}{3}$.
(2nd "Hope Cup" National Mathematics Invitational Competition, 1991)
[Solution] Let the major axis of the ellipse be $2a$, then its minor axis is the diameter of the base circle of the right circular cylinder, which is 2.
The eccentricity of the ellipse is $e=\frac{\sqrt{5}}{3}=\sqrt{1-\left(\frac{1}{a}\right)^{2}}$, yielding $\frac{1}{a}=\frac{2}{3}$. Noting the cosine of the angle $\theta$ between the cross-section and the base of the cylinder,
$$
\cos \theta=\frac{1}{a}, \text { i.e., } \theta=\arccos \frac{2}{3} .
$$

Therefore, the answer is $(C)$.