A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Let AB be the tower.
D is the initial and C is the final position of the car respectively.
BC is the distance from the foot of the tower to the car.

Given,  $\angle ACB={60}^0, \angle ADB={30}^0.$

In $∆ABC,$ we have

$$\begin{gathered} \tan {60}^0=\frac{AB}{BC}. \\ \Rightarrow \sqrt{3}=\frac{AB}{BC}. \\ \Rightarrow BC=\frac{AB}{\sqrt{3}}. \end{gathered}$$

In $∆ABD,$we have

$$\begin{gathered} \tan {30}^0=\frac{AB}{BD}. \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{AB}{BC+CD}. \\ \Rightarrow BC+CD=AB\sqrt{3}. \end{gathered}$$

Substitute the value of BC, then

$$\begin{gathered} CD=AB\sqrt{3}-\frac{AB}{\sqrt{3}}. \\ \Rightarrow CD=AB\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right). \\ \Rightarrow CD=\frac{2AB}{\sqrt{3}}. \end{gathered}$$

Here, distance of BC is half of CD. Thus, the time taken is also half.
Time taken by car to travel distance CD = 6 sec.
Time taken by car to travel BC$=\frac{6}{2}=3$ sec.