In $\mathrm{△}\mathrm{ABC},\mathrm{\angle }\mathrm{B}=\frac{\mathrm{\pi }}{3}\text{ and }\mathrm{\angle }\mathrm{C}=\frac{\mathrm{\pi }}{4}$ Let D divide BC internally in the ratio 1: 3 then $\frac{\sin \mathrm{\angle }\mathrm{BAD}}{\sin \mathrm{\angle }\mathrm{CAD}}=$

Applying sin rule in $∆$ABD

$\frac{\sin \frac{\mathrm{\pi }}{3}}{\mathrm{AD}}=\frac{\sin \mathrm{\angle }\mathrm{BAD}}{\mathrm{BD}}----\left(1\right)$
$\text{ Again from }\mathrm{△}\mathrm{ACD}$
$\begin{array}{l}\frac{\sin \frac{\mathrm{\pi }}{4}}{\mathrm{AD}}=\frac{\sin \mathrm{\angle }\mathrm{CAD}}{\mathrm{DC}}-----\left(2\right)\\ \frac{\left(1\right)}{\left(2\right)}\Rightarrow \frac{\sin \frac{\mathrm{\pi }}{3}}{\sin \frac{\mathrm{\pi }}{4}}=\frac{\mathrm{DC}}{\mathrm{BD}}\frac{\sin \mathrm{\angle }\mathrm{BAD}}{\sin \mathrm{\angle }\mathrm{CAD}}\\ \frac{\sqrt{3}}{2}\times \frac{\sqrt{2}}{1}=\frac{3}{1}\cdot \frac{\sin \mathrm{\angle }\mathrm{BAD}}{\sin \mathrm{\angle }\mathrm{CAD}}\\ \therefore \frac{\sin \mathrm{\angle }\mathrm{BAD}}{\sin \mathrm{\angle }\mathrm{CAD}}=\frac{1}{\sqrt{6}}\end{array}$