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 $30^o$, 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 $60^o$. Find the time taken by the car to reach the foot of the tower from this point.
Given:
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 $30^o$, 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 $60^o$.
To do:
We have to find the time taken by the car to reach the foot of the tower from this point.
Solution:
Let $BCD$ be the highway.
The tower of height $h$ is standing at point $D$.
From the top of the tower of point $A$ the angle of depression is $30^{\circ}$.
After 6 sec when car reaches the point $C$ then angle of depression becomes $60^{\circ}$.
This implies,
The distance covered in $6 \mathrm{sec}=\mathrm{BC}$
From right angled triangle $\mathrm{ADB}$,
$\tan 30^{\circ}=\frac{\mathrm{AD}}{\mathrm{BD}}$
$\frac{1}{\sqrt{3}}=\frac{h}{BD}$
$BD=h \sqrt{3}$.............(i)
From right angled triangle $ADC$,
$\tan 60^{\circ}=\frac{\mathrm{AD}}{\mathrm{CD}}$
$\sqrt{3}=\frac{h}{CD}$
$h=\sqrt{3}CD$.........(ii)
Substituting the value of $h$ in (i), we get,
$BD=\sqrt{3}CD \times \sqrt{3}$
$=3 C D$
$BC+CD=3CD$
$3CD-CD=BC$
$2CD=BC$
$CD=\frac{1}{2} \mathrm{BC}$
The time taken to cover the distance $CD=\frac{1}{2} \times$ Time taken to cover distance $BC$
$=\frac{1}{2} \times 6$
$=3$
Therefore, the time taken by the car to reach the foot of the tower from this point is 3 seconds.
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