The angles of depression of two ships from the top of a light house and on the same side of it are found to be $ 45^{\circ} $ and $ 30^{\circ} $ respectively. If the ships are $ 200 \mathrm{~m} $ apart, find the height of the light house.

Given:

The angles of depression of two ships from the top of a light house and on the same side of it are found to be \( 45^{\circ} \) and \( 30^{\circ} \) respectively.

The ships are \( 200 \mathrm{~m} \) apart.

To do:

We have to find the height of the light house.

Solution:  

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

From the figure,

$\mathrm{CD}=200 \mathrm{~m}, \angle \mathrm{BCA}=30^{\circ}, \angle \mathrm{BDA}=45^{\circ}$

Let the height of the light house be $\mathrm{AB}=h \mathrm{~m}$ and the distance between the first ship $D$ and the light house be $\mathrm{DA}=x \mathrm{~m}$.

This implies,

$\mathrm{CA}=200+x \mathrm{~m}$

We know that,

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

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

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

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

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

Similarly,

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

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

$\Rightarrow \tan 30^{\circ}=\frac{h}{200+x}$

$\Rightarrow \frac{1}{\sqrt3}=\frac{h}{200+h}$              [From (i)]

$\Rightarrow 200+h=h\sqrt3 \mathrm{~m}$

$\Rightarrow 200=h(\sqrt3-1) \mathrm{~m}$

$\Rightarrow 200=h(1.732-1) \mathrm{~m}$

$\Rightarrow h=\frac{200}{0.732} \mathrm{~m}$

$\Rightarrow h=273.2 \mathrm{~m}$

Therefore, the height of the light house is $273.2 \mathrm{~m}$.  

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Updated on: 10-Oct-2022

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