An interesting trigonometric identity I found the following interesting fact:-
$$\dfrac {(\dfrac {\sin 30^0}{\sin 40^0})}{\sin 50^0} = \dfrac {1}{\sin 80^0} \text {(exactly)}$$
It can easily be verified by applying double angle formula.
My questions are:-
1) What is corresponding geometrical figure that can reflect the above; and
2) How can this be applied to give further results [For example, constructing a $50^0$ angle from an $80^0$ one.]?

This is not an answer, but a look at what other solutions to your equation exist.
Your equation can be rewritten as $$\sin(n_1)\sin(n_2) = \sin(n_3)\sin(n_4)$$
I made a small program which checked all combinations of integer angles $n_i$ in the interval [1, 90] degrees. Excluding trivial solutions, where the angles on the LHS are the same as on the RHS, there are $56$ solutions:

Many solutions have the angle $30^{\circ}$. Most of these cases (marked in yellow) have the angles on the other side adding up to $90^{\circ}$. We can deduce the reason for this by seeing that 
$$\begin{align}
\sin(x)\sin(90-x) & = \sin(x)\cos(x) \\
 & = \frac{\sin(2x)}{2} \\ 
 & = \sin(2x)\sin(30^{\circ})
\end{align}$$
The other solutions with angles of $30^{\circ}$, which don't follow the pattern above, are marked in green. I leave it as an exercise for the reader to determine why these, and the unmarked solutions, exist. :)
