Question

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

• A. $2\cos \frac{A}{3}$
• B. $\frac{1}{2}\sec \frac{A}{3}$
• C. $\frac{1}{2}\sin \frac{A}{3}$
• D. $2cosec\frac{A}{3}$

Answer 1

The correct option is B $\frac{1}{2}\sec \frac{A}{3}$

Using Sine law in $\Delta ABD;$

$\frac{AD}{\sin B}=\frac{BD}{\sin A/3}$ (1)

Using Sine law in $\Delta ADC;$

$\frac{AD}{\sin C}=\frac{CD}{\sin 2A/3}$ (2)

divide equation 2 by 1 we get.

$\frac{\sin B}{\sin C}=\frac{\sin A/3}{\sin 2a/3}=\frac{\sin A/3}{2\sin A/3.\cos A/3}=\frac{1}{2\cos A/3}$

Hence;

$\frac{\sin B}{\sin C}=\frac{1}{2}\sec A/3$