[SOLUTION]

We establish a coordinate system where the positive x-axis points East and the positive y-axis points Up. The ground is the stationary frame of reference.

The velocity of the train ($\vec{v}_t$) is west at 10 m/s. In our coordinate system, this is:

$$\vec{v}_t = -10 \hat{i} \text{ m/s}$$

The raindrops have a speed of $v_r = 5$ m/s at an angle $\theta = 30^{\circ}$ west of the vertical. "West of the vertical" means the angle is measured from the negative y-axis towards the negative x-axis. The velocity of the rain ($\vec{v}_r$) has two components: a downward (negative y) component and a westward (negative x) component.

$$v_{rx} = -v_r \sin(\theta)$$ $$v_{ry} = -v_r \cos(\theta)$$

So, the velocity vector of the rain is:

$$\vec{v}_r = (-v_r \sin(\theta)) \hat{i} + (-v_r \cos(\theta)) \hat{j}$$

The velocity of the raindrops relative to the train ($\vec{v}_{r/t}$) is found by vector subtraction:

$$\vec{v}_{r/t} = \vec{v}_r - \vec{v}_t$$

Substituting the vector expressions:

$$\vec{v}_{r/t} = (-v_r \sin(\theta) \hat{i} - v_r \cos(\theta) \hat{j}) - (-10 \hat{i})$$ $$\vec{v}_{r/t} = (10 - v_r \sin(\theta)) \hat{i} - (v_r \cos(\theta)) \hat{j}$$

The speed of the raindrops relative to the train is the magnitude of this vector, $|\vec{v}_{r/t}|$.

$$|\vec{v}_{r/t}| = \sqrt{(10 - v_r \sin(\theta))^2 + (-v_r \cos(\theta))^2}$$

Now, we substitute the given values: $v_r = 5$ m/s and $\theta = 30^{\circ}$.

$$|\vec{v}_{r/t}| = \sqrt{(10 - 5 \sin(30^{\circ}))^2 + (5 \cos(30^{\circ}))^2}$$

Using $\sin(30^{\circ}) = 0.5$ and $\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$:

$$|\vec{v}_{r/t}| = \sqrt{(10 - 5 \cdot 0.5)^2 + (5 \cdot \frac{\sqrt{3}}{2})^2}$$ $$|\vec{v}_{r/t}| = \sqrt{(10 - 2.5)^2 + (2.5\sqrt{3})^2}$$ $$|\vec{v}_{r/t}| = \sqrt{(7.5)^2 + 18.75}$$ $$|\vec{v}_{r/t}| = \sqrt{56.25 + 18.75}$$ $$|\vec{v}_{r/t}| = \sqrt{75} = \sqrt{25 \cdot 3} = 5\sqrt{3}$$

The speed is approximately $8.66$ m/s.