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, A position of man, D be the initial position of car and C be the position of car after 6 seconds.

We know that,

Alternate angles are equal.

From figure,

⇒ ∠ADB = ∠EAD = 30°

⇒ ∠ACB = ∠EAC = 60°

In △ABD,

tan 30° = $\dfrac{\text{Side opposite to angle 30°}}{\text{Side adjacent to angle 30°}}$

Substituting values we get :

$\Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{AB}{BD} \\[1em] \Rightarrow BD = AB\sqrt{3} \text{ m}.$

In △ABC,

tan 60° = $\dfrac{\text{Side opposite to angle 60°}}{\text{Side adjacent to angle 60°}}$

Substituting values we get :

$\Rightarrow \sqrt{3} = \dfrac{AB}{BC} \\[1em] \Rightarrow BC = \dfrac{AB}{\sqrt{3}} \text{ m}.$

From figure,

CD = BD - BC

= $AB\sqrt{3} - \dfrac{AB}{\sqrt{3}} = \dfrac{3AB - AB}{\sqrt{3}} = \dfrac{2AB}{\sqrt{3}}$

= $2 \times \dfrac{AB}{\sqrt{3}} = 2 \times BC$.

According to question,

It takes 6 seconds to cover distance CD or 2BC

and the car is moving with uniform speed.

Since, in 6 seconds car covers a distance of 2BC,

So, a distance of BC meters will be covered in $\dfrac{6}{2}$ = 3 seconds.

Hence, car will take 3 seconds to reach the foot of tower.