6. Given that the base $BC$ and the height $AD$ of isosceles $\triangle ABC$ are both integers. Then, $\sin A$ and $\cos A$ ( ).
(A) one is a rational number, the other is an irrational number
(C) both are rational numbers
(D) both are irrational numbers, which depends on the values of $BC$ and $AD$ to determine
6. B.

Min, $\because S_{\triangle A B C}$
$$
\begin{array}{l}
=\frac{-1}{2} A B \cdot A C \sin A \\
=\frac{1}{2} \cdot B C \cdot A D, \\
\therefore \sin A=\frac{B C \cdot A D}{A B \cdot A C}=\frac{B C \cdot A D}{A B^{2}}=\frac{B C \cdot A D}{B D^{2}+A D^{2}} \\
\quad=\frac{B C \cdot A D}{\frac{1}{4} B C^{2}+A D^{2}}=\frac{4 B C \cdot A D}{B C^{2}+4 A D^{2}}
\end{array}
$$

In $\triangle A B C$, by the cosine rule
$$
\begin{aligned}
\cos A & =\frac{A B^{2}+A C^{2}-B C^{2}}{2 A B \cdot A C}=\frac{2 A B^{2}-B C^{2}}{2 A B^{2}} \\
& =1-\frac{B C^{2}}{2 A B^{2}}=1-\frac{B C^{2}}{2\left(B D^{2}+A D^{2}\right)} \\
& =1-\frac{B C^{2}}{2\left(\frac{1}{4} \cdot B C^{2}+A D^{2}\right)}=1-\frac{2 B C^{2}}{B C^{3}+4 A D^{2}} .
\end{aligned}
$$
$\because B C$ and $A D$ are both integers,
$\therefore \sin A$ and $\cos A$ are both rational numbers.