Question

A man standing on the deck of a ship, which is 10 m above the water level, observes the angle of elevation of the top of a hill as ${60}^{\circ }$, and the angle of depression of the base of the hill as ${30}^{\circ }$. Find the distance of the hill from the ship and the height of the hill.

Answer 1

Let AB be the deck and CD be the hill

Let the man be at B.

Then, AB = 10 m
Let $BE\perp CD$ and $AC\perp CD$

Then, $\angle EBD={60}^{\circ }$ and $\angle EBC={30}^{\circ }$

$\therefore \angle ACB=\angle EBC={30}^{\circ }$

Let CD = h metres

Then, CE = AB = 10 m and

ED $=(h-10)m$

From right $\Delta CAB$, we have

$\Rightarrow \frac{AC}{AB}=\cot {30}^{\circ }$

$\Rightarrow \frac{AC}{10m}=\sqrt{3}$

$\Rightarrow AC=10\sqrt{3}m$

$\therefore BE=AC=10\sqrt{3}m$

From right $\Delta BED$, we have

$\Rightarrow \frac{DE}{BE}=\tan {60}^{\circ }$

$\Rightarrow \frac{h-10}{10\sqrt{3}}=\sqrt{3}$ [using (i)]

$\Rightarrow h-10=30$

$\Rightarrow h=40m$

Hence, the distance of the ship from the hill is $10\sqrt{3}$ metres and the height of the hill is $40m$.