A straight highway leads to the foot of a tower of height 50 m.

Let  be the height of tower m and angle of depression from the top of tower are  and  respectively at two observing Car C and D.

Let m, m and, 

We have the corresponding figure as follows

So we use trigonometric ratios.

In a triangle,

$\Rightarrow \quad \tan D=\frac{A B}{B D}$

$\Rightarrow \quad \tan 30^{\circ}=\frac{50}{x}$

$\Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{50}{x}$

$\Rightarrow \quad x=50 \sqrt{3}$

Since $x=86.6$

Again in a triangle $A B C$

$\Rightarrow \quad \tan C=\frac{A B}{B C}$

$\Rightarrow \quad \tan 60^{\circ}=\frac{50}{x-y}$

$\Rightarrow \quad \sqrt{3}=\frac{50}{x-y}$

$\Rightarrow \sqrt{3} \times 50 \sqrt{3}-\sqrt{3} y=50$

$\Rightarrow \quad y=57.67$

Therefore $x-y=86.6-57.67$

$\Rightarrow \quad x-y=28.93$

Hence the distance of first car from tower is $86.6 \mathrm{~m}$

And the distance of second car from tower is $57.67 \mathrm{~m}$

And the distance between cars is $28.93 \mathrm{~m}$.