Question

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

Answer 1

Let $AB$ be the deck of the ship above the water level and $DE$ be the cliff.
Now,
$AB=16$ m such that $CD=16$ m and ∠​$BDA={30}^o$ and ∠$EBC\hspace{0.17em}={60}^o$.
If AD = x m and $DE=h$ m, then $CE=(h-16)$ m.



In the right ∆$BAD$, we have:
$\frac{AB}{AD}=\tan {30}^o=\frac{1}{\sqrt{3}}$

⇒ $\frac{16}{x}=\frac{1}{\sqrt{3}}$

⇒ $x=16\sqrt{3}$ $=27.68$ m

In the right ∆$EBC$, we have:
$\frac{EC}{BC}=\tan {60}^o=\sqrt{3}$

⇒ $\frac{(h-16)}{x}=\sqrt{3}$
⇒ $h-16=x\sqrt{3}$
⇒ $h-16=16\sqrt{3}\times \sqrt{3}=48$ [∵ $x=16\sqrt{3}$]
⇒ $h=48+16=64$ m

∴ Distance of the cliff from the deck of the ship = $AD=x=27.68$ m
And,
Height of the cliff = $DE=h=64$ m