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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
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Verified Answer
The correct answer is:
$\frac{1}{\sqrt{6}}$
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
$$
\begin{aligned}
& \frac{\sqrt{3} x}{2 \sin (\angle \mathrm{BAD})}=\frac{3 x}{\sqrt{2} \sin (\angle \mathrm{CAD})} \\
\therefore \quad & \frac{\sin (\angle \mathrm{BAD})}{\sin (\angle \mathrm{CAD})}=\frac{\sqrt{6}}{6}=\frac{1}{\sqrt{6}}
\end{aligned}
$$
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