Let the ground be the stationary frame of reference. We define a coordinate system where the positive x-direction ($\hat{i}$) is forward (the direction of train motion) and the positive y-direction ($\hat{j}$) is vertically downward.

The velocity of the rain relative to the ground is $\vec{v}_{rg} = v_x \hat{i} + v_y \hat{j}$, where $v_x$ is the horizontal component (wind) and $v_y$ is the vertical component. The velocity of the train relative to the ground is $\vec{v}_{t}$. The velocity of the rain relative to the train, which determines the angle of the streaks, is $\vec{v}_{rt} = \vec{v}_{rg} - \vec{v}_{t}$. The angle $\theta$ of the streaks from the vertical is given by $\tan\theta = \frac{\text{horizontal component of } \vec{v}_{rt}}{\text{vertical component of } \vec{v}_{rt}}$.

Case 0: Train is stopped The train's velocity is $\vec{v}_{t} = \vec{0}$.

$\vec{v}_{rt} = \vec{v}_{rg} = v_x \hat{i} + v_y \hat{j}$.

The streaks are tilted forward at $\theta_0$, meaning $v_x$ is in the positive direction.

$$ \tan\theta_0 = \frac{v_x}{v_y} \quad (1) $$

Case 1: Train moves at velocity $v_1$ The train's velocity is $\vec{v}_{t} = v_1 \hat{i}$.

$\vec{v}_{rt,1} = (v_x - v_1)\hat{i} + v_y \hat{j}$.

The streaks are tilted backward at $\theta_1$, meaning the net horizontal component is negative ($v_x - v_1 < 0$).

$$ \tan\theta_1 = \frac{|v_x - v_1|}{v_y} = \frac{v_1 - v_x}{v_y} \quad (2) $$

Case 2: Train moves at velocity $v_2$ The train's velocity is $\vec{v}_{t} = v_2 \hat{i}$.

$\vec{v}_{rt,2} = (v_x - v_2)\hat{i} + v_y \hat{j}$.

The streaks are tilted backward at $\theta_2$, meaning the net horizontal component is negative ($v_x - v_2 < 0$).

$$ \tan\theta_2 = \frac{|v_x - v_2|}{v_y} = \frac{v_2 - v_x}{v_y} \quad (3) $$

To find the ratio $v_1/v_2$, we solve this system of equations. From (1), express the horizontal wind speed as $v_x = v_y \tan\theta_0$. Substitute $v_x$ into equation (2):

$$ \tan\theta_1 = \frac{v_1 - v_y \tan\theta_0}{v_y} = \frac{v_1}{v_y} - \tan\theta_0 $$

Solving for $v_1$:

$$ v_1 = v_y (\tan\theta_0 + \tan\theta_1) $$

Substitute $v_x$ into equation (3):

$$ \tan\theta_2 = \frac{v_2 - v_y \tan\theta_0}{v_y} = \frac{v_2}{v_y} - \tan\theta_0 $$

Solving for $v_2$:

$$ v_2 = v_y (\tan\theta_0 + \tan\theta_2) $$

Finally, compute the ratio $v_1/v_2$:

$$ \frac{v_1}{v_2} = \frac{v_y (\tan\theta_0 + \tan\theta_1)}{v_y (\tan\theta_0 + \tan\theta_2)} $$