A) \[\frac{\pi }{6}\]
B) \[\frac{\pi }{3}\]
C) \[\frac{\pi }{2}\]
D) \[\pi \]
Correct Answer: A
Solution :
We have \[a+b+c=\frac{6(\sin A+\sin B+\sin C)}{3}\] Þ \[k(\sin A+\sin B+\sin C)=2(\sin A+\sin B+\sin C)\], where \[k=\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\] Þ \[k=2\], \[(\because \sin A+\sin B+\sin C\ne 0)\] \[\therefore \] \[\frac{a}{\sin A}=2\] Þ \[\sin A=\frac{1}{2}\], (\[\because \] \[a=1\]) Þ \[A=\frac{\pi }{6}\].You need to login to perform this action.
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