Assume north to be $\hat{\mathrm{i}}$ direction and vertically downward to be -$\hat{\mathrm{j}}$.
Let the rain velocity, $\overrightarrow{\mathrm{v}}$ = a$\hat{\mathrm{i}}$ + b$\hat{\mathrm{j}}$.
$\overrightarrow{{\mathrm{v}}_{\mathrm{r}}} = \mathrm{a}\hat{\mathrm{i}}+\mathrm{b}\hat{\mathrm{j}}$
 Given velocity of girl = $\overrightarrow{{\mathrm{v}}_{\mathrm{g}}}$ = (5 m/s)$\hat{\mathrm{i}}$

Let $\overrightarrow{{\mathrm{v}}_{\mathrm{rg}}}$ = Velocity of rain w.r.t girl

$\begin{array}{l}= \overrightarrow{{\mathrm{v}}_{\mathrm{r}}}-\overrightarrow{{\mathrm{v}}_{\mathrm{g}}} = (\mathrm{a}\hat{\mathrm{i}}+\mathrm{b}\hat{\mathrm{j}}) - 5\hat{\mathrm{i}}\\ = (\mathrm{a}-5)\widehat{\hat{\mathrm{i}} }+ \mathrm{b}\hat{\mathrm{j}}\end{array}$

According to the question rain, appears to fall vertically downward.
Hence,             a - 5 = 0 $\Rightarrow$ a = 5

Step 2: Find the vertical velocity of rain.

Given the velocity of the girl, $\overrightarrow{{\mathrm{v}}_{\mathrm{g}}}$ = (10 m/s)$\hat{\mathrm{i}}$
$\begin{array}{l}\overrightarrow{{\mathrm{v}}_{\mathrm{rg}}}=\overrightarrow{{\mathrm{v}}_{\mathrm{r}}}-\overrightarrow{{\mathrm{v}}_{\mathrm{g}}}\\ =(a\hat{i}+b\hat{j})-10\hat{i}=(a-10)\hat{i}+b\hat{j}\end{array}$

According to question rain appears to fall at 45 to the vertical hence
$\Rightarrow$ b = a -10 = 5 - 10 = - 5
Hence, velocity of rain = a$\hat{\mathrm{i}}$ + b$\hat{\mathrm{j}}$

$\begin{array}{l}\Rightarrow {\mathrm{v}}_{\mathrm{r}}=5\hat{\mathrm{i}}-5\hat{\mathrm{j}}\\ \mathbf{Step}\mathbf{ }\mathbf{3}\mathbf{:} \mathrm{Find} \mathrm{speed} \mathrm{of} \mathrm{rain}, \mathrm{and} \mathrm{direction} \mathrm{of} \mathrm{rain} \mathrm{fall} \mathrm{w}.\mathrm{r}.\mathrm{to} \mathrm{earth} \mathrm{frame}\\ \text{ Speed of rain }=\left|{\mathrm{v}}_{\mathrm{r}}\right|=\sqrt{(5{)}^2+(-5{)}^2}=\sqrt{50}=5\sqrt{2}\mathrm{m}/\mathrm{s}\\ \tan \theta = \frac{-5}{5} = -1\\ \theta = - {45}^{°}.\\ \end{array}$