Question

If in a $△ABC,\angle B=\frac{\pi }{3}$ and $\angle C=\frac{\pi }{4},$ let $D$ divide $BC$ internally in the ratio $1:3$.Then $\frac{\sin \left(\angle BAD\right)}{\sin \left(\angle CAD\right)}$ is equal to

• A. $\frac{1}{\sqrt{6}}$
• B. $\frac{1}{3}$
• C. $\frac{1}{\sqrt{3}}$
• D. $\sqrt{\frac{2}{3}}$

Answer 1

The correct option is A $\frac{1}{\sqrt{6}}$

$\because BD:DC=1:3$
$\therefore BD=\frac{a}{4}$ and $DC=\frac{3a}{4}$
In $△ABD,$ sine rule and $△ADC$ we have
$\frac{AD}{\sin \frac{\pi }{3}}=\frac{\frac{a}{4}}{\sin \alpha }$ and
$\frac{\frac{3a}{4}}{\sin \beta }=\frac{AD}{\sin \frac{\pi }{4}}$
$\therefore AD.\sin \alpha =\frac{a}{4}\sin \frac{\pi }{3}=\frac{a}{4}\frac{\sqrt{3}}{2}=\frac{a\sqrt{3}}{8}$ ................$\left(1\right)$
$AD.\sin \beta =\frac{3a}{4}\sin \frac{\pi }{4}=\frac{3a}{4\sqrt{2}}$ ............$\left(2\right)$
Dividing $\left(1\right)$ by $\left(2\right)$ we get
$\frac{\sin \alpha }{\cos \alpha }=\frac{a\sqrt{3}}{8}\times \frac{4\sqrt{2}}{3a}=\frac{1}{\sqrt{6}}$