Question : In a $\triangle ABC, \angle B=\frac{\pi}{3}, \angle C=\frac{\pi}{4}$ and D divides BC internally in the ratio 1 : 3, then $\frac{\sin \angle BAD}{\sin \angle CAD}$ is equal to:
Option 1: $\frac{1}{\sqrt{2}}$
Option 2:
$\frac{1}{\sqrt{3}}$
Option 3:
$\frac{1}{\sqrt{6}}$
Option 4: $\sqrt{6}$
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Correct Answer:
$\frac{1}{\sqrt{6}}$
Solution :
Given:
$\triangle ABC, \angle B=\frac{\pi}{3}=\frac{180°}{3}=60°,$ and $\angle C=\frac{\pi}{4}=\frac{180°}{4}=45°$
So, $\angle A= 180°-(60°+45°)=75°$
In triangles $ABD$ and $ACD$, we have:
$\frac{AB}{\sin B}=\frac{BD}{\sin \angle BAD}$ and
$\frac{AD}{\sin C}=\frac{CD}{\sin \angle CAD}$
⇒ $\frac{\sin C}{\sin B}=\frac{\sin \angle CAD}{\sin \angle BAD}×\frac{BD}{CD}$
⇒ $\frac{\sin 60°}{\sin 45°}=\frac{\sin \angle CAD}{\sin \angle BAD}×\frac{1}{3}$
⇒ $\frac{\sin \angle BAD}{\sin \angle CAD}=\frac{1}{3}×\frac{\frac{\sqrt{3}}{2}}{\frac{1}{\sqrt{2}}}=\frac{1}{\sqrt{6}}$
Hence, the correct answer is $\frac{1}{\sqrt{6}}$.
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