As observed from the top of a $ 150 \mathrm{~m} $ tall light house, the angles of depression of two ships approaching it are $ 30^{\circ} $ and $ 45^{\circ} $. If one ship is directly behind the other, find the distance between the two ships.

Given:

As observed from the top of a \( 150 \mathrm{~m} \) tall light house, the angles of depression of two ships approaching it are \( 30^{\circ} \) and \( 45^{\circ} \).

One ship is exactly behind the other on the same side of the lighthouse.

To do:

We have to find the distance between the two ships.

Solution:  

Let $AB$ be the height of the tall lighthouse and $C, D$ are the two ships one behind the other.

From the figure,

$\mathrm{AB}=150 \mathrm{~m}, \angle \mathrm{BCA}=30^{\circ}, \angle \mathrm{BDA}=45^{\circ}$

Let the distance between ship $D$ and the lighthouse be $\mathrm{DA}=x \mathrm{~m}$ and the distance between the two ships be $\mathrm{CD}=y \mathrm{~m}$.

This implies,

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

We know that,

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

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

$\Rightarrow \tan 45^{\circ}=\frac{75}{x}$

$\Rightarrow 1=\frac{150}{x}$

$\Rightarrow x=150 \mathrm{~m}$.........(i)

Similarly,

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

$=\frac{\text { BA }}{CA}$

$\Rightarrow \tan 30^{\circ}=\frac{150}{x+y}$

$\Rightarrow \frac{1}{\sqrt3}=\frac{150}{150+y}$              [From (i)]

$\Rightarrow 150+y=150\sqrt3 \mathrm{~m}$

$\Rightarrow y=150(\sqrt3-1) \mathrm{~m}$            

$\Rightarrow y=150(1.73-1) \mathrm{~m}$ 

$\Rightarrow y=150(0.73) \mathrm{~m}$ 

$\Rightarrow y=109.5 \mathrm{~m}$ 

Therefore, the distance between the two ships is $109.5 \mathrm{~m}$.  

Tutorialspoint

e

Updated on: 10-Oct-2022

21 Views

Kickstart Your Career

Get certified by completing the course

Get Started