A man standing on the deck of a ship, which is 8 m above water level.

Solution:

Let H be height of hill CE and a man is standing on a ships at the height of 8meter above from the water level. Let AB = 8, BC = x, AD = BC, AB = DC, DE = h. , and 

We have to find x and H

The corresponding figure is as follows

In $\triangle A B C$,

$\Rightarrow \quad \tan 30=\frac{8}{x}$

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

$\Rightarrow \quad x=8 \sqrt{3}$

Again in $\triangle D A E$,

$\Rightarrow \quad \tan 60^{\circ}=\frac{h}{x}$

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

$\Rightarrow \quad h=x \sqrt{3}$

$\Rightarrow \quad h=24$

\text { Therefore } H=h+8

$\Rightarrow \quad H=24+8$

$\Rightarrow \quad H=32$

Hence the required distance is $8 \sqrt{3} \mathrm{~m}$ and height is $32 \mathrm{~m}$.