A) \[{{90}^{o}}\]
B) \[{{45}^{o}}\]
C) \[{{120}^{o}}\]
D) None of these
Correct Answer: C
Solution :
Since \[\cos 3A+\cos 3B+\cos 3C=1\] Þ \[4\sin \frac{3A}{2}\sin \frac{3B}{2}\sin \frac{3C}{2}=0\] Either \[\frac{3A}{2}={{180}^{o}}\]or \[\frac{3B}{2}={{180}^{o}}\]or \[\frac{3C}{2}={{180}^{o}}\] Either \[A={{120}^{o}}\]or \[B={{120}^{o}}\]or \[C={{120}^{o}}\].
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