Question

In $△ABC,$ the median AD divides $\angle BAC$ such that $\angle BAC:\angle CAD=2:1.$ Then $\cos \frac{A}{3}$ 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 $\frac{\sin B}{2\sin C}$

Consider the below given triangle.

Applying sine rule to triangle $BAD$ we get

$\frac{a}{2sin\frac{2A}{3}}=\frac{AD}{sinB}$.
Or
$\frac{a.sinB}{2sin\frac{2A}{3}}=AD$.

And
Applying sine rule to triangle $CAD$

$\frac{a.sinC}{2sin\frac{A}{3}}=AD$

Hence
$\frac{a.sinC}{2sin\frac{A}{3}}=\frac{a.sinB}{2sin\frac{2A}{3}}$
Or
$sinC.sin\frac{2A}{3}=sinB.sin\frac{A}{3}$.
Or
$sinC[2sin\frac{A}{3}.cos\frac{A}{3}]=sinB.sin\frac{A}{3}$
Or
$2sinC.cos\frac{A}{3}=sinB$
Or
$cos\frac{A}{3}=\frac{sinB}{2sinC}$.