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:
![](/assets/questions/media/158630-44487-1620147857.jpg)
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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