A man on the deck of a ship is 10 m above the water level. He observes that the angle of elevation of the top of a cliff is 45° and the angle of depression of the base is 30°. Calculate the distance of the cliff from the ship and the height of the cliff.
Let H be the height of the cliff CE. And a man is standing on the ships at the height of
10 meter above from the water level. Let AB = 10, BC = x, AD = BC, AB = DC, DE = h. and
We have to find H and x
The corresponding figure is as follows
In $\triangle A B C$,
$\Rightarrow \quad \tan C=\frac{A B}{B C}$
$\Rightarrow \quad \tan 30^{\circ}=\frac{10}{x}$
$\Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{10}{x}$
$\Rightarrow \quad x=10 \sqrt{3}$
Again in $\triangle D A E$,
$\Rightarrow \quad \tan A=\frac{D E}{A D}$
$\Rightarrow \quad \tan 45^{\circ}=\frac{h}{x}$
$\Rightarrow \quad \mathrm{I}=\frac{h}{x}$
$\Rightarrow \quad x=h$
$\Rightarrow x=10 \sqrt{3}$
Therefore $H=h+10$
$\Rightarrow \quad H=10 \sqrt{3}+10$
$\Rightarrow \quad H=10(\sqrt{3}+1)$
$\Rightarrow \quad H=27.32$
Hence the required distance is $10 \sqrt{3} \mathrm{~m}$. and height is $27.32 \mathrm{~m}$.
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