In a triangle ABC, with the usual notations, | vedclass

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(C) Let $BD = x$ and $DC = 3x$, so $BC = 4x$. In $	riangle ABC$, by the Sine Rule, $\frac{AB}{\sin C} = \frac{AC}{\sin B} \implies \frac{AB}{AC} = \f

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

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(C) Let $BD = x$ and $DC = 3x$, so $BC = 4x$. In $	riangle ABC$, by the Sine Rule, $\frac{AB}{\sin C} = \frac{AC}{\sin B} \implies \frac{AB}{AC} = \frac{\sin(\pi/4)}{\sin(\pi/3)} = \frac{1/\sqrt{2}}{\sqrt{3}/2} = \sqrt{\frac{2}{3}}$.In $	riangle ABD$, by the Sine Rule, $\frac{BD}{\sin \angle BAD} = \frac{AB}{\sin \angle ADB} \implies \sin \angle BAD = \frac{BD \cdot \sin \angle ADB}{AB}$.In $	riangle ACD$, by the Sine Rule, $\frac{DC}{\sin \angle CAD} = \frac{AC}{\sin \angle ADC} \implies \sin \angle CAD = \frac{DC \cdot \sin \angle ADC}{AC}$.Since $\angle ADB + \angle ADC = \pi$, $\sin \angle ADB = \sin \angle ADC$. Thus, $\frac{\sin \angle BAD}{\sin \angle CAD} = \frac{BD}{AB} \cdot \frac{AC}{DC} = \frac{x}{AB} \cdot \frac{AC}{3x} = \frac{1}{3} \cdot \frac{AC}{AB}$.Substituting $\frac{AC}{AB} = \sqrt{\frac{3}{2}}$, we get $\frac{\sin \angle BAD}{\sin \angle CAD} = \frac{1}{3} \cdot \sqrt{\frac{3}{2}} = \frac{1}{\sqrt{3} \cdot \sqrt{2}} = \frac{1}{\sqrt{6}}$.

In a triangle $ABC$, with the usual notations, $\angle B = \frac{\pi}{3}$ and $\angle C = \frac{\pi}{4}$. If $D$ divides $BC$ internally in the ratio $1:3$, then $\frac{\sin \angle BAD}{\sin \angle CAD} =$

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