A man on the deck of a ship is $ 10 \mathrm{~m} $ above the water level. He observes that the angle of elevation of the top of a cliff is $ 45^{\circ} $ and the angle of depression of the base is $ 30^{\circ} $. Calculate the distance of the cliff from the ship and the height of the cliff.

Given:

A man on the deck of a ship is \( 10 \mathrm{~m} \) above the water level. He observes that the angle of elevation of the top of a cliff is \( 45^{\circ} \) and the angle of depression of the base is \( 30^{\circ} \).

To do:

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

Solution:

Let $CD$ be the cliff 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 $45^{o}$.

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

From the figure,

$\angle ADB =30^{o}, AB=10\ m$ and  $\angle CBE=45^{o}$

This implies,

$ED=AB=10\ m$ and $CE=h-10\ m$

In $\vartriangle CBE$,

$tan 45^{o} =\frac{CE}{BE} =\frac{h-10}{x}$

$1 =\frac{h-10}{x}$

$x=h-10$.........(i)

In $\vartriangle ABD$,

$tan 30^{o}=\frac{AB}{AD} =\frac{10}{x}$

$\frac{1}{\sqrt{3}} =\frac{10}{x}$

$x=10\sqrt{3}\ m$

Substituting $x=10\sqrt{3}$ in equation (i), we get,

$10\sqrt{3}=h-10$

$h=10+10\sqrt{3}\ m$

$h=10(1.732+1)\ m$

$h=10(2.732) = 27.32\ m$

$\Rightarrow x=27.32-10=17.32\ m$

Therefore, the distance of the cliff from the ship is $17.32 \ m$ and the height of the cliff is $27.32\ m$.

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Updated on: 10-Oct-2022

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