From the top of a $ 120 \mathrm{~m} $ high tower, a man observes two cars on the opposite sides of the tower and in straight line with the base of tower with angles of depression as $ 60^{\circ} $ and $ 45^{\circ} $. Find the distance between the cars. (Take $ \sqrt{3}=1.732 $ ).

Given:

From the top of a \( 120 \mathrm{~m} \) high tower, a man observes two cars on the opposite sides of the tower and in straight line with the base of tower with angles of depression as \( 60^{\circ} \) and \( 45^{\circ} \).

To do:

We have to find the distance between the cars.

Solution:  

Let $AB$ be the height of the tower and $C, D$ be the points where the two cars are whose angles of depression are \( 45^{\circ} \) and \( 60^{\circ} \) respectively.

From the figure,

$\mathrm{AB}=120 \mathrm{~m}, \angle \mathrm{BCA}=60^{\circ}, \angle \mathrm{BDA}=45^{\circ}$     

Let the car $C$ be at a distance of $\mathrm{AC}=x \mathrm{~m}$ from the foot of the tower and the distance between the car $D$ and the foot of the tower be $\mathrm{AD}=y \mathrm{~m}$.

This implies,

$\mathrm{CD}=x+y \mathrm{~m}$

We know that,

$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$

$=\frac{\text { AB }}{DA}$

$\Rightarrow \tan 45^{\circ}=\frac{120}{y}$

$\Rightarrow 1=\frac{120}{y}$

$\Rightarrow y=120 \mathrm{~m}$..........(i)

Similarly,

$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$

$=\frac{\text { AB }}{AC}$

$\Rightarrow \tan 60^{\circ}=\frac{120}{x}$

$\Rightarrow \sqrt3=\frac{120}{x}$                        

$\Rightarrow x=\frac{120}{\sqrt3} \mathrm{~m}$

$\Rightarrow x=\frac{40\sqrt3(\sqrt3)}{\sqrt3} \mathrm{~m}$           

$\Rightarrow x=40(1.732) \mathrm{~m}$

$\Rightarrow x=69.28 \mathrm{~m}$........(ii)

From (i) and (ii), we get,

$x+y=120+69.28 \mathrm{~m}$

$\Rightarrow x+y=189.28 \mathrm{~m}$

Therefore, the distance between the cars is $189.28 \mathrm{~m}$.

Tutorialspoint

e

Updated on: 10-Oct-2022

745 Views

Kickstart Your Career

Get certified by completing the course

Get Started