Question

If in $△ABC,\sum \sin 3A=0$, then at least one angle of $△ABC$ is

• A. ${60}^0$
• B. ${30}^0$
• C. ${90}^0$
• D. ${45}^0$

Answer 1

The correct option is A ${60}^0$

Given $\sum sin3A=0$

We know $sin3A=3sinA-4si{n}^{3}A$

$sin3A=0$

$3sinA-4si{n}^{3}A=0$

$\left(sin3A\right)(3-4si{n}^{2}A)=0$

$sinA=0$ and $(3-4si{n}^{2}A)=0$

In a triangle $sinA$ cannot be zero

So $(3-4si{n}^{2}A)=0$

$sinA=\frac{\sqrt{3}}{2}$

$A={60}^0$