Question

A man is standing on the deck of a ship, which is $10m$ above water level. He observes the angle of elevation of the top of a light house as ${60}^0$ and the angle of depression of the base of lighthouse as ${30}^0$. Find the height of the light house.

• A. $40m$
• B. $50m$
• C. $60m$
• D. $30m$

Answer 1

The correct option is A $40m$

Let $CD$ be the lighthouse and suppose the man is standing on the deck of a ship at point $A$

The angle of depression of the base $C$ of the lighthouse $CD$ observed from $A$ is ${30}^{\circ }$ and the angle of elevation of the top $D$ of the lighthouse $CD$ observed from $A$ is ${60}^{\circ }$

$\therefore \angle EAD={60}^{\circ }$ and $\angle BCA={30}^{\circ }$

In $△AED,$

$\tan {60}^{\circ }=\frac{DE}{EA}$

$\therefore \sqrt{3}=\frac{h}{x}$

$\therefore h=\sqrt{3}x$ ..........$\left(1\right)$

In $△ABC,$

$\tan {30}^{\circ }=\frac{AB}{BC}$

$\therefore \frac{1}{\sqrt{3}}=\frac{10}{x}$

$\therefore x=10\sqrt{3}$

Substituting $x=10\sqrt{3}$ in equation $\left(1\right)$ we get

$h=\sqrt{3}x=\sqrt{3}\times 10\sqrt{3}=30$

$\therefore DE=30m$

$\therefore CD=CE+ED=(10+30)m=40m$

Thus, the height of the lighthouse is $40m$