Directional Dilemma

We see that the the rain has three components, two horizontal and one vertical. If we let these components be $x$ , $y$ , and $z$ , $z$ being the horizontal component, then if we face North, we will see components $z$ and $x$ , while if we face East we will see components $z$ and $y$ ,

From the components we see if we face North we can draw the conclusion that $z = x tan( 60^{\circ} )$ .

In a similar way, from the components we see if we face East we can draw the conclusion that $z = y tan( 30^{\circ} )$ .

Since we want to know from which direction the rain originated, we have to resort to top view, where we can only see components $x$ and $y$ . Component $x$ is towards East, and Component $y$ is towards North.

To find the angle $\theta$ the rain creates from east, we use the fact that $tan$ $\theta = \frac {y}{x}$ .

That becomes

$\huge tan (\theta) = \frac { \frac {z}{tan( 30^{\circ} )} }{ \frac {z}{tan( 60^{\circ} )}}$

$\large tan (\theta) = \frac{tan( 60^{\circ}) } { tan(30^{\circ}) }$

$tan (\theta) = 3$ .

That roughly gives $\theta = 71.565^{\circ}$ North of East. This directly means, too, that the wind must come somewhere South West.

Since we are to determine the origin of the wind that causes this rainfall to move in such direction, we subtract $180^{\circ}$ from it (because this will be the direction from which the wind comes from) and we get $-108.435^{\circ}$ or roughly $\boxed{-108^{\circ}}$ . This basically means that the wind comes from $108.435^{\circ}$ South (Negative of North) of East.

Forgive me if there were no illustrations in the solution/problem. But I think brilliant minds would be able to imagine this easy problem. :) Please don't hesitate to reshare!