The rain is falling along $O A$ with speed $30 \mathrm{~ms}^{-1}$ and woman rider is moving along $O S$ with $10 \mathrm{~ms}^{-1}$, i.e. $O A=30 \mathrm{~ms}^{-1}$ and $O B=10 \mathrm{~ms}^{-1}$.


The woman rider can protect herself from the rain if she holds her umbrella in the direction of relative velocity of rain w.r.t. woman. To do so apply equal and opposite velocity of woman on the rain i.e. impress the velocity $10 \mathrm{~ms}^{-1}$ due north on rain which is represented by $O C$. Now the relative velocity of rain w.r.t. woman will be represented by diagonal $O D$ of parallelogram $O A D C$.
If $\angle A O D=\theta$, then in $\triangle O A D$,
$\tan \theta=A D / O A=O C / O A=10 / 30=0.3333$ $=\tan 18^{\circ} 26^{\prime}$
$\beta=18^{\circ} 26^{\prime}$ with vertical in forward direction.