Question

If the median $AD$ of a triangle $ABC$ divides the angle $\angle BAC$ in the ratio $1:2$, then $\frac{\sin \left(B\right)}{\sin \left(C\right)}$ 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. $2\csc \left(\frac{A}{3}\right)$

Answer 1

The correct option is B

$\frac{1}{2}\sec \left(\frac{A}{3}\right)$

Explanation of the correct option.

Compute the required value.

From sine rule in $∆ABD$,

$\frac{AD}{\sin \left(B\right)}=\frac{BD}{\sin \left({\displaystyle \frac{A}{3}}\right)}$……………$\left(1\right)$

From sine rule in $∆ADC$,

$\frac{AD}{\sin \left(C\right)}=\frac{CD}{\sin \left({\displaystyle \frac{2A}{3}}\right)}$…………$\left(2\right)$

Divide equation $\left(2\right)$ by equation $\left(1\right)$,

$\begin{array}{rcl}\frac{\sin \left(B\right)}{\sin \left(C\right)}& =& \frac{\sin \left({\displaystyle \frac{A}{3}}\right)}{\sin \left({\displaystyle \frac{2A}{3}}\right)}\\ & \Rightarrow & \frac{\sin \left(B\right)}{\sin \left(C\right)}=\frac{\sin \left({\displaystyle \frac{A}{3}}\right)}{2\sin \left({\displaystyle \frac{A}{3}}\right)\cos \left({\displaystyle \frac{A}{3}}\right)}\\ & \Rightarrow & \frac{\sin \left(B\right)}{\sin \left(C\right)}=\frac{1}{2\cos \left({\displaystyle \frac{A}{3}}\right)}\\ & \Rightarrow & \frac{\sin \left(B\right)}{\sin \left(C\right)}=\frac{1}{2}\sec \left(\frac{A}{3}\right)\end{array}$

Hence option $\left(B\right)$ is the correct option.