Question

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

Answer 1

On the basis of the given information, consider the figure shown above.

Let $AB=d$

In $△AB{P}_1,\angle BA{P}_1=60°$

We know that, $\tan \theta =\frac{\text{Opposite Side}}{\text{Adjacent Side}}$

$\therefore \tan 60°=\frac{{BP}_1}{d}$

$\Rightarrow {BP}_1=\sqrt{3}d[\because \tan 60°=\sqrt{3}]$

Similarly, in $△AB{P}_2,\angle BA{P}_2=30°$

$\therefore \tan 30°=\frac{{BP}_2}{d}$

$\Rightarrow {BP}_2=\frac{d}{\sqrt{3}}[\because \tan 30°=\frac{1}{\sqrt{3}}]$

Now, ${P_{1}P}_2={BP}_1-B{P}_2$

$=d\sqrt{3}-\frac{d}{\sqrt{3}}$

$=\frac{2d}{\sqrt{3}}$

So, the car took $6$ seconds to travel $\frac{2d}{\sqrt{3}}$ distance.

So, to reach foot of tower the car should travel the distance $B{P}_2$

We know that, $speed=\frac{distance}{time}$

$\therefore Speed=\frac{\frac{2d}{\sqrt{3}}}{6}$

$=\frac{d}{3\sqrt{3}}$

Hence, to travel $B{P}_2$ distance

$time=\frac{distance}{speed}$

$=\frac{\frac{d}{\sqrt{3}}}{\frac{d}{3\sqrt{3}}}$

$=3$ seconds

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