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If in a triangle ABC , $s i n 3 A + \sin 3 B + \sin 3 C = 0$ then atleast one of the angle is

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$\frac{π}{3}$

$\frac{π}{6}$

$\frac{5 π}{6}$

$\frac{π}{2}$

answer is B.

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Detailed Solution

$\sin 3 A + \sin 3 B + \sin 3 C$

$= 4 \cos \frac{3 A}{2} . \cos \frac{3 B}{2} . \cos \frac{3 C}{2} = 0$

$\frac{3 A}{2} = \frac{π}{2} o r \frac{3 B}{2} = \frac{π}{2} o r \frac{3 C}{2} = \frac{π}{2}$

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