A man running on a horizontal road at $8km/h$ finds the rain falling vertically. He increases his speed to 12 km/h and finds that the drops make angle 30⁰ with the vertical. Find the speed of the rain with respect to the road.

We have  ${\overrightarrow{\upsilon }}_{rain,road}={\overrightarrow{\upsilon }}_{rain,man}+{\overrightarrow{\upsilon }}_{man,road}\mathrm{......}\left(1\right)$
The two situations given in the problem may be represented by the following figure.
 
${\overrightarrow{\upsilon }}_{rain,road}$  is same in magnitude and direction in both the figures.
Taking horizontal components in equation (i) for figure
  ${\overrightarrow{\upsilon }}_{rain,road}\sin \alpha =8km/h.\mathrm{....}\left(ii\right)$
Now consider figure. Draw a line $OA\perp {\overrightarrow{\upsilon }}_{rain,man}$ as shown.
Taking components in equation (i) along the line OA, 
 ${\overrightarrow{\upsilon }}_{rain,road}\sin (30°+\alpha )=12km/h\cos 30°.\mathrm{.....}\left(iii\right)$
From (ii) and (iii),
 $\frac{\sin (30°+\alpha )}{\sin \alpha }=\frac{12\times \sqrt{3}}{8\times 2}$
or,  $\frac{\sin 30°\cos \alpha +\cos 30°\sin \alpha }{\sin \alpha }=\frac{3\sqrt{3}}{4}$
or,  $\frac{1}{2}\cot \alpha +\frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{4}$
or,  $\cot \alpha =\frac{\sqrt{3}}{2}$
or,  $\alpha ={\cot }^{-1}\frac{\sqrt{3}}{2}$
From (ii),  ${\upsilon }_{rain,road}=\frac{8km/h}{\sin \alpha }=4\sqrt{7}km/h$