The figure from below shows three masses tied to a central one which is hanging with respect of a floor. Find the angle so that the system remains in static equilibrium:

The alternatives are as follows:

$\begin{array}{ll}

1.&50^{\circ}\\

2.&80^{\circ}\\

3.&60^{\circ}\\

4.&70^{\circ}\\

5.&45^{\circ}\\

\end{array}$

What I've attempted to do was to equate what it is on the vertical components for the tension of the string and for the horizontal components.

What I obtained was as follows:

In the vertical:

$Mg\cos 50^{\circ}+mg\cos \alpha = mg$

In the horizontal:

$Mg\sin 50^{\circ} = mg\sin\alpha$

Therefore:

$M=\frac{m\sin\alpha}{\sin 50^{\circ}}$

If I do insert this in the equation for the vertical components. I'm obtaining the following:

$\frac{mg\sin\alpha}{\sin 50^{\circ}}\cos 50^{\circ}+mg\cos \alpha = mg$

Dividing by similar terms and multiplying by \sin 50^{\circ} to tall:

$\sin\alpha\cos 50^{\circ}+\cos \alpha\sin50^{\circ}=\sin 50$

This is reduced to:

$\sin(\alpha+50^{\circ})=\sin 50^{\circ}$

Therefore $\alpha =0$

But this doesn't make sense. Could it be that I'm missinterpreting something. Can somebody help me here?.