Question

If the median $AD$ of a triangle $ABC$ divides the $\angle BAC$ in the ratio $1:2,$ then $\frac{\sin B}{\sin C}$ is equal to:

• A. $2\cos \left(\frac{A}{3}\right)$
• B. $\frac{1}{2}\sec \left(\frac{A}{3}\right)$
• C. $\frac{1}{2}\sin \left(\frac{A}{3}\right)$
• D. $2\text{cosec}\left(\frac{A}{3}\right)$

Answer 1

The correct option is B $\frac{1}{2}\sec \left(\frac{A}{3}\right)$

$\frac{\sin \left(\frac{A}{3}\right)}{\frac{a}{2}}=\frac{\sin B}{AD}$ and $\frac{\sin \frac{2A}{3}}{\frac{a}{2}}=\frac{\sin C}{AD}$ ...... [Sine rule in $△ABD$ and $△ACD$]

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

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

$\frac{\sin B}{\sin C}=\frac{1}{2}\sec \left(\frac{A}{3}\right)$

Hence, $\frac{1}{2}\sec \left(\frac{A}{3}\right)$ is the correct answer.