In a triangle ABC, $\mathrm{\angle }\mathrm{B}=\frac{\mathrm{\pi }}{3},\mathrm{\angle }\mathrm{C}=\frac{\mathrm{\pi }}{4}$ and D divides BC internally in the ratio 1:3. Then $\frac{\sin \mathrm{\angle }\mathrm{BAD}}{\sin \mathrm{\angle }\mathrm{CAD}}$ is equal to

$$\begin{gathered} In ∆BAD u\sin g \sin e rule \\ \frac{\sin \angle \mathrm{BAD}}{1}=\frac{\sin {60}^{\circ }}{\mathrm{AD}}\Rightarrow AD=\frac{\sqrt{3}}{2\left(\sin \angle \mathrm{BAD}\right)} \\ \text{ and }In ∆CAD u\sin g \sin e rule\frac{\sin \mathrm{\angle }\mathrm{CAD}}{3}=\frac{\sin {45}^{\circ }}{\mathrm{AD}}\Rightarrow AD=\frac{3}{\sqrt{2}\sin \mathrm{\angle }\mathrm{CAD}} \\ \Rightarrow \frac{\sqrt{3}}{2\left(\sin \angle \mathrm{BAD}\right)}\frac{3}{\sqrt{2}\sin \mathrm{\angle }\mathrm{CAD}}\Rightarrow \frac{\left(\sin \angle \mathrm{BAD}\right)}{\sin \mathrm{\angle }\mathrm{CAD}}=\frac{\sqrt{6}}{6}=\frac{1}{\sqrt{6}} \end{gathered}$$