Question

If $a\sin \alpha =b\sin {(120}^o+\alpha )=c\sin {(240}^o+\alpha )$, prove that, $bc+ca+ab=0$.

Answer 1

Consider the given equation.

$a\sin \alpha =b\sin (120{}^{\circ }+\alpha )=c\sin (240{}^{\circ }+\alpha )=k$ (say)

Therefore,

$\frac{k}{a}=\sin \alpha$

$\frac{k}{b}=\sin (120{}^{\circ }+\alpha )$

$\frac{k}{c}=\sin (240{}^{\circ }+\alpha )=\sin (x-120{}^{\circ })$

So,

$k(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=(\sin \alpha +\sin (120{}^{\circ }+\alpha )+\sin (\alpha -120{}^{\circ }))$

$=\sin \alpha +\sin \alpha \cos 120{}^{\circ }+\cos \alpha \sin 120{}^{\circ }+\sin \alpha cos120{}^{\circ }-cos\alpha sin120{}^{\circ }$

$=sin\alpha +2\sin \alpha \cos 120{}^{\circ }$

$=\sin \alpha +2(-\frac{1}{2})\sin \alpha$

$=0$

As, $k\ne 0$,

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$

$ab+bc+ca=0$

Hence, proved.