Question

In a $△ABC,$ if $\sum \sin 3A=0,$ then it is

• A. equilateral
• B. right angled
• C. isosceles
• D. has at least one angle ${60}^0$

Answer 1

Answer: (A), (D)

The correct options are
A equilateral
D has at least one angle ${60}^0$

$\sum \sin 3A=0$
$\Rightarrow \sin 3A+\sin 3B+\sin 3C=0$ by expanding the summation
$\Rightarrow 2\sin \left(\frac{3}{2}(A+B)\right)\cos \left(\frac{3}{2}(A-B)\right)+\sin 3C=0$ using compound angle formula.
$\Rightarrow 2\sin \left(\frac{3}{2}(A+B)\right)\cos \left(\frac{3}{2}(A-B)\right)+2\sin \frac{3C}{2}\cos \frac{3C}{2}=0$ using multiple angle formula.
$\Rightarrow 2\sin (\frac{3\pi }{2}-\frac{3C}{2})\cos \left(\frac{3}{2}(A-B)\right)+2\sin \frac{3C}{2}\cos \frac{3C}{2}=0$ since $A+B+C=\pi$
$\Rightarrow -2\cos \frac{3C}{2}\cos \left(\frac{3}{2}(A-B)\right)+2\sin \frac{3C}{2}\cos \frac{3C}{2}=0$
$\Rightarrow -\cos \frac{3C}{2}[\cos \left(\frac{3}{2}(A-B)\right)-\cos \frac{3C}{2}]=0$ by taking common terms
$\Rightarrow \cos \frac{3C}{2}[\cos \left(\frac{3}{2}(A-B)\right)+\cos \left(\frac{3}{2}(A+B)\right)]=0$ since $\cos (\pi -\theta )=-\cos \theta$
$\Rightarrow \cos \frac{3C}{2}\cos \frac{3A}{2}\cos \frac{3B}{2}=0$ by compound angle formula
$\Rightarrow \cos \frac{3C}{2}=\cos {90}^0$
$\Rightarrow \frac{3C}{2}={90}^0$
$\Rightarrow C={60}^0$
$\therefore C={60}^0$ or $A={60}^0$ or $B={60}^0$