4. Given a regular tetrahedron $S-ABC$, the base is an equilateral triangle with side length 1, and the side edges are 2. If a section through the line $AB$ divides the volume of the regular tetrahedron into two equal parts, then the cosine of the plane angle of the dihedral angle formed by the section and the base is
A. $\frac{\sqrt{15}}{10}$
B. $\frac{4 \sqrt{15}}{15}$
C. $\frac{\sqrt{15}}{15}$
D. $\frac{2 \sqrt{15}}{15}$
Solution
As shown in the figure, let $M, D$ be the midpoints of $SC, AB$, respectively. Then the section $MAB$ is the one we are looking for, and $\angle MDC$ is the plane angle of the dihedral angle formed by the section and the base.
$AM$ is the median of $\triangle ASC \Rightarrow 2(AM^2 + MC^2) = AS^2 + AC^2$ $\Rightarrow AM^2 = \frac{5}{2} - 1 = \frac{3}{2} \Rightarrow MD^2 = \frac{3}{2} - \frac{1}{4} = \frac{5}{4}$,
Thus, in $\triangle MDC$, $\cos \angle MDC = \frac{\frac{5}{4} + \frac{3}{4} - 1}{2 \cdot \frac{\sqrt{5}}{2} \cdot \frac{\sqrt{3}}{2}} = \frac{2 \sqrt{15}}{15}$.
Therefore, the answer is $D$.