In a Delta ABC,angle B = frac{pi}{3},angle C | vedclass

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(C) Given: $\angle B = \frac{\pi}{3}$,$\angle C = \frac{\pi}{4}$ and $\frac{BD}{DC} = \frac{1}{3}$.In $\Delta ABD$,by the Law of Sines:$\frac{BD}{\sin

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$\frac{1}{\sqrt{6}}$

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(C) Given: $\angle B = \frac{\pi}{3}$,$\angle C = \frac{\pi}{4}$ and $\frac{BD}{DC} = \frac{1}{3}$.In $\Delta ABD$,by the Law of Sines:$\frac{BD}{\sin \angle BAD} = \frac{AD}{\sin \angle B}$$\Rightarrow \frac{BD}{\sin \angle BAD} = \frac{AD}{\sin(\pi/3)} = \frac{AD}{\sqrt{3}/2}$$\Rightarrow AD = \frac{\sqrt{3}}{2} \cdot \frac{BD}{\sin \angle BAD}$ .... $(1)$In $\Delta ADC$,by the Law of Sines:$\frac{DC}{\sin \angle CAD} = \frac{AD}{\sin \angle C}$$\Rightarrow \frac{DC}{\sin \angle CAD} = \frac{AD}{\sin(\pi/4)} = \frac{AD}{1/\sqrt{2}}$$\Rightarrow AD = \frac{1}{\sqrt{2}} \cdot \frac{DC}{\sin \angle CAD}$ .... $(2)$Equating $(1)$ and $(2)$:$\frac{\sqrt{3}}{2} \cdot \frac{BD}{\sin \angle BAD} = \frac{1}{\sqrt{2}} \cdot \frac{DC}{\sin \angle CAD}$$\Rightarrow \frac{\sin \angle BAD}{\sin \angle CAD} = \frac{\sqrt{3}/2}{1/\sqrt{2}} \cdot \frac{BD}{DC}$$\Rightarrow \frac{\sin \angle BAD}{\sin \angle CAD} = \frac{\sqrt{3}}{2} \cdot \sqrt{2} \cdot \frac{1}{3} = \frac{\sqrt{3}}{\sqrt{2} \cdot 3} = \frac{1}{\sqrt{2} \cdot \sqrt{3}} = \frac{1}{\sqrt{6}}$.

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

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