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

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(C) Let $\angle BAD = \alpha$ and $\angle CAD = \beta$.In $\Delta ADB$,applying the sine rule:$\frac{BD}{\sin \alpha} = \frac{AD}{\sin B} \implies \fr

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $\angle BAD = \alpha$ and $\angle CAD = \beta$.In $\Delta ADB$,applying the sine rule:$\frac{BD}{\sin \alpha} = \frac{AD}{\sin B} \implies \frac{x}{\sin \alpha} = \frac{AD}{\sin(\pi/3)}$ .....$(i)$In $\Delta ADC$,applying the sine rule:$\frac{CD}{\sin \beta} = \frac{AD}{\sin C} \implies \frac{3x}{\sin \beta} = \frac{AD}{\sin(\pi/4)}$ .....(ii)Dividing $(i)$ by (ii):$\frac{x}{\sin \alpha} 	imes \frac{\sin \beta}{3x} = \frac{AD}{\sin(\pi/3)} 	imes \frac{\sin(\pi/4)}{AD}$$\frac{\sin \beta}{3 \sin \alpha} = \frac{1/\sqrt{2}}{\sqrt{3}/2} = \frac{2}{\sqrt{6}} = \sqrt{\frac{2}{3}}$$\frac{\sin \beta}{\sin \alpha} = 3 	imes \sqrt{\frac{2}{3}} = \sqrt{3} 	imes \sqrt{2} = \sqrt{6}$Therefore,$\frac{\sin \angle BAD}{\sin \angle CAD} = \frac{\sin \alpha}{\sin \beta} = \frac{1}{\sqrt{6}}$.

In a triangle $ABC$,$\angle B = \frac{\pi}{3}$ and $\angle C = \frac{\pi}{4}$,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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