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

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(C) Given $\cos 3A + \cos 3B + \cos 3C + \cos 3\pi = 0$. Since $\cos 3\pi = -1$,we have $\cos 3A + \cos 3B + \cos 3C = 1$. Using the identity $\cos 3A

Vedclass · Accepted Answer

$\frac{\pi}{3}$

Vedclass · Answer

(C) Given $\cos 3A + \cos 3B + \cos 3C + \cos 3\pi = 0$. Since $\cos 3\pi = -1$,we have $\cos 3A + \cos 3B + \cos 3C = 1$. Using the identity $\cos 3A + \cos 3B + \cos 3C = 1 - 4 \cos \frac{3A}{2} \cos \frac{3B}{2} \cos \frac{3C}{2}$ for $A+B+C = \pi$,we get: $1 - 4 \cos \frac{3A}{2} \cos \frac{3B}{2} \cos \frac{3C}{2} = 1$ $\Rightarrow 4 \cos \frac{3A}{2} \cos \frac{3B}{2} \cos \frac{3C}{2} = 0$. This implies $\cos \frac{3A}{2} = 0$ or $\cos \frac{3B}{2} = 0$ or $\cos \frac{3C}{2} = 0$. For $	riangle ABC$,$0 $\cos \frac{3A}{2} = 0$ $\Rightarrow \frac{3A}{2} = \frac{\pi}{2}$ $\Rightarrow A = \frac{\pi}{3}$. If one angle is $\frac{\pi}{3}$,the sum of the other two angles is $\pi - \frac{\pi}{3} = \frac{2\pi}{3}$. However,if we consider the case where one angle is $\frac{2\pi}{3}$,then the sum of the other two is $\frac{\pi}{3}$. Thus,the least value of the sum of two angles is $\frac{\pi}{3}$.

List-$I$	List-$II$
$(A)$ $\sum \cot A$	$(i)$ $\frac{(a+b+c)^2}{4\Delta}$
$(B)$ $\sum \cot \frac{A}{2}$	$(ii)$ $\frac{a^2+b^2+c^2}{4\Delta}$
$(C)$ If $\tan A : \tan B : \tan C = 1 : 2 : 3$,then $\sin A : \sin B : \sin C =$	$(iii)$ $8 : 6 : 5$
$(D)$ If $\cot \frac{A}{2} : \cot \frac{B}{2} : \cot \frac{C}{2} = 3 : 7 : 9$,then $a : b : c =$	$(iv)$ $12 : 5 : 13$
	$(v)$ $\sqrt{5} : 2\sqrt{2} : 3$
	$(vi)$ $4\Delta$

In $\triangle ABC$,if $\cos 3A + \cos 3B + \cos 3C + \cos 3\pi = 0$,then the least value of the sum of two of its angles is

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