Question

In a triangle $ABC,\angle =\pi /3$ and $\angle C=\pi /4$. Let $D$ divide $BC$ internally in the ratio $1:3$. Then $\frac{\sin \angle BAD}{\sin \angle CAD}$ equals.

• 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 $\frac{1}{\sqrt{6}}$

In triangles $ABD$ and $ACD$, we have

$\frac{AB}{\sin B}=\frac{BD}{\sin \angle BAD}$

and $\frac{AD}{\sin C}=\frac{CD}{\sin \angle CAD}$

$\Rightarrow \frac{\sin C}{\sin B}=\frac{\sin \angle CAD}{\sin \angle BAD}.\frac{BD}{CD}$

$\Rightarrow \frac{\sin \frac{\pi }{4}}{\sin \frac{\pi }{3}}=\frac{\sin \angle CAD}{\sin \angle BAD}.\frac{1}{3}$

$\Rightarrow \frac{\sin \angle BAD}{\sin \angle CAD}=\frac{1}{3}\times \frac{\sqrt{3}/2}{1/\sqrt{2}}=\frac{1}{\sqrt{6}}$