A man on the deck of a ship, $16m$ above water level observes that the angles of elevation and depression respectively of the top and bottom of a cliff are ${60}^{\circ }$ and ${30}^{\circ }.$ Calculate the distance of the cliff from the ship and height of the cliff.

Let $AB$ be the deck of the ship above the water level and $DE$ be the cliff.
Now,
$AB=16m$ such that $CD=16m$ and $\angle BDA={30}^{\circ }$ and $\angle EBC={60}^{\circ }$.
If $AD=xm$ and $DE=hm,$ then $CE=(h-16)m.$

In the right $△BAD,$ we have:
$\frac{AB}{AD}=\tan {30}^{\circ }=\frac{1}{\sqrt{3}}$
$\Rightarrow \frac{16}{x}=\frac{1}{\sqrt{3}}$
$\Rightarrow x=16\sqrt{3}=27.68m$
In the right $△EBC,$ we have:
$\frac{EC}{BC}=\tan {60}^{\circ }=\sqrt{3}$
$\Rightarrow \frac{(h-16)}{x}=\sqrt{3}$
$\Rightarrow h-16=x\times \sqrt{3}$
$\Rightarrow h-16=16\sqrt{3}\times \sqrt{3}=48$ $[\because x=16\sqrt{3}]$
$\Rightarrow h=48+16=64m$
$\therefore$ Distance of the cliff from the deck of the ship $=AD=x=27.68m$
And,
Height of the cliff $=DE=h=64m$