2. In acute $\triangle A B C$, $B E$ and $C F$ are altitudes on sides $A C$ and $A B$, respectively. Then $S_{\triangle A E F}: S_{\triangle A B C}=(\quad$.
(A) $\sin ^{2} A$
(B) $\cos ^{2} A$
(C) $\operatorname{tg}^{2} \hat{A}$
(D) ctg $A$
2. B.

In the right triangle $\triangle A B E$,
$$
\begin{aligned}
\cos A & =\frac{A E}{A B}, \text { in the right triangle } \triangle A C F, \\
\cos A & =\frac{A F}{A C}, \\
& \therefore \frac{S_{\triangle A E F}}{S_{\triangle A B C}} \\
& =\frac{\frac{1}{2} A E \cdot A F \sin A}{\frac{1}{2} A B \cdot A C \sin A} \\
& =\frac{A E}{A B} \cdot \frac{A F}{A C}=\cos ^{2} A .
\end{aligned}
$$