Relative Velocity Problem: Finding Train Speed from Raindrop Angle

[Mr Davis 97]

Homework Statement

A person looking out the window of a stationary train notices that rain drops are falling vertically down at a speed of 5.0 m/s relative to the ground. When the train moves at a constant velocity, the raindrops make an angle of 25° when they move past the window. How fast is the train moving?

Homework Equations

$\vec{v_{AC}} = \vec{v_{AB}} + \vec{v_{BC}}$

The Attempt at a Solution

T = train, R = rain, G = ground
$\vec{v_{TG}} = \vec{v_{TR}} + \vec{v_{RG}}$
$\vec{v_{TG}} = \vec{v_{RG}} - \vec{v_{RT}}$
$v_x = 0~m/s-[(5~m/s)\sin25^{\circ}] = 2.1131~m/s$
$v_y = (-5~m/s) -[(5~m/s)\cos25^{\circ}] = -0.468~m/s$
$\vec{v_{TG}} = \sqrt{(2.1131~m/s)^2 + (-0.468~m/s)^2} = 2.16~m/s$

However, according to the solution manual, the correct answer is 2.3 m/s. Where am I going wrong?

 

[RJLiberator]

Try drawing a right triangle.

The vertical length is the trains speed.
The horizontal length is the rain drop speed of 5.0.

Which angle are you looking from?
What trig will help you solve it in one quick calculation?

 

[Mr Davis 97]

Try drawing a right triangle.

The vertical length is the trains speed.
The horizontal length is the rain drop speed of 5.0.

Which angle are you looking from?
What trig will help you solve it in one quick calculation?

What do you mean "the vertical length is the train's speed"? Do you mean horizontal?