Obtaining the absolute velocity of a moving train based on angle of raindrops with respect to vertical axis

To find the angle $\theta$ we need to find the vertical speed and the horizental speed  of the raindrop with respect to the moving train. Note that the velocity vector of the raindrop is given by 
\[(U_{wind} \sin (\theta_{wind}),-U_{wind}\cos(\theta_{wind})).\]
The speed of the train in vector form is
\[(V_{train}, 0 ).\]
Note that if the train is moving to the right we assume $V_{train}>0$ and if it is moving to the left then $V_{train}<0$. 

The vertical speed of the raindrop with respect to the moving  train is 
\[S_{vertical}=U_{wind}\cos (\theta_{wind}),\]
and the raindrop is moving downwords. 
The horizental speed of the raindrop with respect to the moving train is
\[S_{horizental}=V_{train}+U_{wind} \sin (\theta_{wind}).\]
Then 
\[\tan(\theta)=\frac{S_{horizental}}{S_{vertical}}.\]
Thus
\[\theta= \arctan( \frac{V_{train}+U_{wind} \sin (\theta_{wind})}{U_{wind}\cos (\theta_{wind})} ).\]