In a triangle ABC,cos 3A + cos 3B + cos 3C = | vedclass

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(C) Given $\cos 3A + \cos 3B + \cos 3C = 1$. Using the identity for $\cos 3A + \cos 3B + \cos 3C$ in a triangle where $A+B+C = \pi$,the expression sim

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$3$

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(C) Given $\cos 3A + \cos 3B + \cos 3C = 1$. Using the identity for $\cos 3A + \cos 3B + \cos 3C$ in a triangle where $A+B+C = \pi$,the expression simplifies to $1 - 4 \cos \frac{3A}{2} \cos \frac{3B}{2} \cos \frac{3C}{2} = 1$. This implies $\cos \frac{3A}{2} \cos \frac{3B}{2} \cos \frac{3C}{2} = 0$. Thus,one of the angles must satisfy $\frac{3A}{2} = \frac{\pi}{2}$,$\frac{3B}{2} = \frac{\pi}{2}$,or $\frac{3C}{2} = \frac{\pi}{2}$. This gives $A = \frac{\pi}{3}$,$B = \frac{\pi}{3}$,or $C = \frac{\pi}{3}$. However,for the sum to be $1$,the condition leads to one angle being $120^\circ$ (i.e.,$\frac{2\pi}{3}$). Let $A = \frac{2\pi}{3}$. Given circumradius $R = \sqrt{3}$. Using the sine rule,the side $a = 2R \sin A = 2(\sqrt{3}) \sin \frac{2\pi}{3} = 2\sqrt{3} 	imes \frac{\sqrt{3}}{2} = 3$. Since $A$ is the largest angle $(120^\circ)$,$a$ is the longest side. Therefore,the length of the longest side is $3$.

In a triangle $ABC$,$\cos 3A + \cos 3B + \cos 3C = 1$. If the circumradius of triangle $ABC$ is $\sqrt{3}$,then the length of its longest side is

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