Question

In a $\Delta ABC$, $\angle B=\frac{\pi }{3}$ and $\angle C=\frac{\pi }{4}$. If point D divides BC internally in the ratio $1:3$, then $\frac{\sin \angle BAD}{\sin \angle CAD}$ is equal to

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

Answer 1

Answer: (A)

$1/\sqrt{6}$

Applying sine rule in $△ABD$, we get $\frac{p}{\sin \angle BAD}=\frac{AD}{\sin {60}^o}$....(1)
Applying sine rule in $△CAD$, we get $\frac{3p}{\sin \angle CAD}=\frac{AD}{\sin {45}^o}$....(2)
Dividing (2) by (1), we get $\frac{3\sin \angle BAD}{\sin \angle CAD}=\frac{\sin {60}^o}{\sin {45}^o}$
.
$\Rightarrow \frac{\sin \angle BAD}{\sin \angle CAD}=\frac{1}{3}\times \frac{\sqrt{3}}{2}\times \sqrt{2}=\frac{1}{\sqrt{6}}$