Question

In $\Delta ABC,$ the median $AD$ divides $\angle BAC$ such that $\angle BAD:\angle CAD=2:1$.

Then cos $\left(\frac{A}{3}\right)$ is equal to?

• A. $\frac{\sin B}{2\sin C}$
• B. $\frac{\sin C}{2\sin B}$
• C. $\frac{2\sin B}{\sin C}$
• D. none of these

Answer 1

The correct option is A $\frac{\sin B}{2\sin C}$

By sine rule in $△ABD,\frac{\sin \frac{2A}{3}}{\frac{a}{2}}=\frac{\sin B}{AD}--------\left(1\right)$

By sine rule in $△ADC,\frac{\sin \frac{A}{3}}{\frac{a}{2}}=\frac{\sin C}{AD}---------\left(2\right)\frac{eq\left(1\right)}{eq\left(2\right)}=\frac{2\sin \frac{A}{3}\cos \frac{A}{3}}{\sin \frac{A}{3}}=\frac{\sin B}{\sin C}$

$\cos \frac{A}{3}=\frac{\sin B}{2\sin C}$

Hence, $\frac{\sin B}{2\sin C}$ is the correct answer.