In \(\triangle \mathrm{ABC}, \mathrm{m} \angle \mathrm{B}=\frac{\pi}{3}\) and \(\mathrm{m} \angle \mathrm{C}=\frac{\pi}{4}\). Let point \(\mathrm{D}\) divide \(\mathrm{BC}\) internally in the ratio \(1: 3\), then \(\frac{\sin (\angle B A D)}{\sin (\angle C A D)}\) has the value

In \(\triangle \mathrm{ABD}\),
\(
\begin{aligned}
& \frac{\sin (\angle \mathrm{BAD})}{\mathrm{BD}}=\frac{\sin (\angle \mathrm{ABD})}{\mathrm{AD}} \\
& \Rightarrow \frac{\sin (\angle \mathrm{BAD})}{x}=\frac{\frac{\sqrt{3}}{2}}{\mathrm{AD}} \\
& \Rightarrow \mathrm{AD}=\frac{\sqrt{3} x}{2 \sin (\angle \mathrm{BAD})}
\end{aligned}
\)
In \(\triangle \mathrm{ADC}\),
\(
\begin{aligned}
& \frac{\sin (\angle \mathrm{CAD})}{\mathrm{DC}}=\frac{\sin (\angle \mathrm{ACD})}{\mathrm{AD}} \\
\Rightarrow & \frac{\sin (\angle \mathrm{CAD})}{3 x}=\frac{\frac{1}{\sqrt{2}}}{\mathrm{AD}} \\
\therefore \quad & \mathrm{AD}=\frac{3 x}{\sqrt{2} \sin (\angle \mathrm{CAD})}
\end{aligned}
\)
From (i) and (ii), we get
\(
\begin{aligned}
& \frac{\sqrt{3} x}{2 \sin (\angle \mathrm{BAD})}=\frac{3 x}{\sqrt{2} \sin (\angle \mathrm{CAD})} \\
\therefore \quad & \frac{\sin (\angle \mathrm{BAD})}{\sin (\angle \mathrm{CAD})}=\frac{\sqrt{6}}{6}=\frac{1}{\sqrt{6}}
\end{aligned}
\)