Question

Question 15
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 ${30}^{\circ }$, 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 ${60}^{\circ }$. Find the time taken by the car to reach the foot of the tower from this point.

Answer 1

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

According to question,
In right $\Delta ABC,$
tan ${60}^{\circ }=\frac{AB}{BC}$
$\Rightarrow \sqrt{3}=\frac{AB}{BC}$
$\Rightarrow BC=\frac{AB}{\sqrt{3}}$

Also,
In right $\Delta ABD$,
tan ${30}^{\circ }=\frac{AB}{BD}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{AB}{(BC+CD)}$
$\Rightarrow AB\sqrt{3}=BC+CD$
$\Rightarrow AB\sqrt{3}=\frac{AB}{\sqrt{3}}+CD$
$\Rightarrow CD=AB\sqrt{3}-\frac{AB}{\sqrt{3}}$
$\Rightarrow CD=AB(\sqrt{3}-\frac{1}{\sqrt{3}})$
$\Rightarrow CD=\frac{2AB}{\sqrt{3}}$

$\Rightarrow AB=\frac{\sqrt{3}CD}{2}$

Substitute this AB value in BC

$\Rightarrow BC=\frac{\sqrt{3}CD}{2\sqrt{3}}$

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

Time taken by car to travel $BC=\frac{6}{2}=3sec.$

Hence,the time taken by car to reach the foot of the tower from the given point is $3sec.$