A man standing on the deck of a ship, which is $10\ m$ above water level, observes the angle of elevation of the top of a hill as $60^{o}$ and the angle of depression of the base of a hill as $30^{o}$. Find the distance of the hill from the ship and the height of the hill.

Given: Height of the deck from the water level$=10\ m$, angle of the elevation of the top of the hill$=60^{o}$ and the angle of depression of the base of a hill as $30^{o}$. 

To do: To Find the distance of the hill from the ship and the height of the hill.

Solution:

Let CD be the hill and suppose the man is standing on the deck of a ship at point A.

The angle of depression of the base C of the hill CD observed from A is $30^{o}$

And the angle of elevation of the top D of the hill CD observed from A is $60^{o}$.

$\therefore  \angle EAD =60^{o}$ and  $\angle BCA=30^{o}$

In $\vartriangle AED$,

$tan 60^{o} =\frac{DE}{EA} =\frac{h}{x}$

$\sqrt{3} =\frac{h}{x}$

$h=\sqrt{3} x\ \ ...............( 1)$

In $\vartriangle ABC$,

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

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

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

Substituting, in equation $( 1)$ , we get $x=10\sqrt{3}$

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

$=10 \times \ 3\ =\ 30\ m$

$DE = 30\ m$

$CD=CE + ED = 10 + 30 = 40\ m$

Thus, The distance of the hill from the ship is $10\sqrt{3} \ m$ and the height of the hill is $40\ m$.

 

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Updated on: 10-Oct-2022

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