Question

A man on the deck of a ship, $14$m above the water level, observes that the angle of elevation of the top of a cliff is ${60}^{\circ }$ and angle of depression of the base of the cliff is ${30}^{\circ }$. Find the height of the cliff.

Answer 1

Let $BD$ be the height of the cliff.

Let $A$ be the position of the ship and $E$ be the point of observation so that $AE=14$m.

Draw $EC$ parallel to $AB$ such that $AB=EC$.

Given that $\angle ABE={30}^{\circ },\angle DEC={60}^{\circ }$

In the right angled $△ABE,\tan {30}^{\circ }=\frac{AE}{AB}$

$\therefore AB=\frac{AE}{\tan {30}^{\circ }}$

$\Rightarrow AB=14\sqrt{3}$

Thus, $EC=14\sqrt{3}$

In the right angled $△DEC,\tan {60}^{\circ }=\frac{CD}{EC}$

$\therefore CD=EC\tan {60}^{\circ }$

$\Rightarrow CD=\left(14\sqrt{3}\right)\sqrt{3}=42$ m

Thus, the height of the cliff,

$BCD=BC+CD$

$=14+42$

$=56$ m.