question_answer

In \[\Delta ABC,\] the median divides \[\angle BAC\] such that \[\angle BAD:\angle CAD=2:1.\] Then the value of \[\cos \left( \frac{A}{3} \right)\]equals

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

Correct Answer: A

Solution :

[a] Using sine rule in \[\Delta ABD\] and \[\Delta ADC,\] we get \[\frac{AD}{\sin B}=\frac{BD}{\sin \frac{2A}{3}}\]  and  \[\frac{AD}{\sin \,\,C}=\frac{CD}{\sin \frac{A}{3}}\] \[\therefore \,\,\,\frac{\sin B}{\sin C}=\frac{\sin \frac{2A}{3}}{\sin \frac{A}{3}}\]   \[(\because \,BD=CD)\] \[\therefore \,\,\,\,\cos \frac{A}{3}=\frac{\sin B}{2\sin C}\]