Is this Question correct? Newtonian Motion - Relative Motion of Rain

Question

There is a question in a textbook which states:

"A cyclist is riding north at 12km/h when it starts to rain. The rain appears to be falling towards her at an angle of 10 degrees relative to the vertical. Deciding to return home, the cyclist turns south, riding at the same speed. Now the rain appears to be coming towards her at an angle of 6 degrees to the vertical. What is the velocity of the rain?"

Now, the answers say that the velocity of the rain is 3.5km/h at an angle of 27 degrees to the vertical.

However, I don't see any situation is which this is possible. By travelling north, you're already extending the horizontal velocity of the rain which places it at 10 degrees. By travelling south, you've reduced the velocity of the rain to 0 and then reversed the horizontal velocity so that it now travels at 6 degrees relative to you. Therefore the angle of the rain when you are stationary shouldn't exceed 6 degrees or 10 degrees? I feel as if the interpretation of the question I've used may not be correct.

Answer 1

$\let\a=\alpha$ I believe I've found where the error is. Solver exchanged axes when relating angles to velocity components. The problem, elementary as it may appear, is rather tricky about axes orientation, component signs, angles definitions.

I'll take an $x$-axis northwards, a $y$-axis upwards (this choice makes $y$-component of rain negative, but helps with angles). Initial cyclist's velocity is $u_1=12\,\rm km/h$ (positive $x$ direction). Rain's velocity may have both components, $v_x$ and $v_y$, unknown.

In cyclist's frame rain's velocity has components $v'_x=v_x-u_1$, $v'_y=v_y$. In this frame rain is seen falling at an angle $\a'=10^\circ$. It obeys $$\tan\a_1 = v'_x/v'_y.\tag1$$ Here is the solver's error: he/she wrote (I suppose) $$\tan\a_1 = v'_y/v'_x \tag{wrong!}$$ which would be true if $\a'$ were the angle formed with $x$-axis.

From (1) $${v_x - u_1 \over v_y} = \tan\a_1.\tag2$$

Now for backward ride. Cyclist's velocity has become $u_2=-u_1$. Rain's velocity wrt ground is the same as before, but cyclist has changed her motion, so rain's relative velocity has a different $x$-component: $v''_x=v_x-u_2$, $v''_y=v_y$. As to rain's angle $\a''=-6^\circ$ (negative because rain is seen to come from negative $x$-direction) we have $$\tan\a_2 = v''_x/v''_y$$ $${v_x - u_2 \over v_y} = \tan\a_2.\tag3$$

Eqs. (2), (3) are two linear equations for the two unknowns $v_x$, $v_y$. Solution is

$$v_x = {u_1 \tan\a_2 - u_2 \tan\a_1 \over \tan\a_2 - \tan\a_1} = -3.04\,\rm km/h$$ $$v_y = {u_1 - u_2 \over \tan\a_2 - \tan\a_1} = -85.3\,\rm km/h.$$ $$v = \sqrt{v_x^2 + v_y^2} = 85.3\,\rm km/h$$ $$\a = \arctan{v_x \over v_y} = 2.04^\circ.$$

(Please check my solution, especially numerical results.)