A man standing on the deck of a ship, which is $ 8 \mathrm{~m} $ above water level. He observes the angle of elevation of the top of a hill as $ 60^{\circ} $ and the angle of depression of the base of the hill as $ 30^{\circ} $. Calculate the distance of the hill from the ship and the height of the hill.

Given:

A man standing on the deck of a ship, which is \( 8 \mathrm{~m} \) above water level. He observes the angle of elevation of the top of a hill as \( 60^{\circ} \) and the angle of depression of the base of the hill as \( 30^{\circ} \).

To do:

We have to find the distance of the hill from the ship and the height of the hill.

Solution:

Let $CD$ be the hill and the man is standing on the deck of a ship $AB$ at point $B$.

The angle of depression of the base $D$ of the cliff $CD$ observed from $B$ is $30^{o}$ and the angle of elevation of the top $C$ of the cliff $CD$ observed from $B$ is $60^{o}$.

Let the height of the cliff be $h\ m$.

From the figure,

$\angle ADB =30^{o}, AB=8\ m$ and  $\angle CBE=60^{o}$

This implies,

$ED=AB=8\ m$ and $CE=h-8\ m$

In $\vartriangle CBE$,

$tan 60^{o} =\frac{CE}{BE} =\frac{h-8}{x}$

$\sqrt3=\frac{h-8}{x}$

$x=\frac{h-8}{\sqrt3}$.........(i)

In $\vartriangle ABD$,

$tan 30^{o}=\frac{AB}{AD} =\frac{8}{x}$

$\frac{1}{\sqrt{3}} =\frac{8}{x}$

$x=8\sqrt{3}\ m$............(ii)

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

$\frac{h-8}{\sqrt3}=8\sqrt3$

$h-8=8\sqrt{3}(\sqrt3)\ m$

$h-8=8(3)\ m$

$h=24+8 = 32\ m$

Therefore, the distance of the hill from the ship is $8\sqrt3\ m$ and the height of the hill is $32\ m$. 

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Updated on: 10-Oct-2022

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