1. In $\triangle A B C$, $A B=A C$. If $\sin A=\frac{3}{5}$, then $\frac{B C}{A B}=$ . $\qquad$
$=.1 . \frac{\sqrt{10}}{5}$.
As shown in Figure 7, draw $BH \perp AC$, with $H$ as the foot of the perpendicular.
$$
\begin{array}{l}
\text { Let } BH=3k, \\
AB=AC=5k .
\end{array}
$$

Then $AH=4k$,
$$
\begin{array}{c}
CH=k, \\
BC=\sqrt{BH^{2}+CH^{2}} \\
=\sqrt{10} k .
\end{array}
$$

Therefore, $\frac{BC}{AB}=\frac{\sqrt{10}}{5}$.