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 \[\frac{AD}{\sin \,\,C}=\frac{CD}{\sin \frac{A}{3}}\] \[\therefore \,\,\,\frac{\sin B}{\sin C}=\frac{\sin \frac{2A}{3}}{\sin \frac{A}{3}}\] \[(\because \,BD=CD)\] \[\therefore \,\,\,\,\cos \frac{A}{3}=\frac{\sin B}{2\sin C}\]You need to login to perform this action.
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