In triangle ABC,m angle B = frac{pi}{3} and m | vedclass

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(C) In $	riangle ABD$,by the Law of Sines:$\frac{\sin(\angle BAD)}{BD} = \frac{\sin(\angle B)}{AD}$$\Rightarrow \frac{\sin(\angle BAD)}{x} = \frac{\s

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

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

In $\triangle ABC$,$m \angle B = \frac{\pi}{3}$ and $m \angle C = \frac{\pi}{4}$. Let point $D$ divide $BC$ internally in the ratio $1:3$. Then,the value of $\frac{\sin(\angle BAD)}{\sin(\angle CAD)}$ is

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