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. [3 MARKS]

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

$\frac{AC}{AB}=cot{30}^{\circ }=\sqrt{3}\Rightarrow \frac{AC}{10m}=\sqrt{3}$

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

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

From right $\Delta BED$, we have

$\frac{DE}{BE}=tan{60}^{\circ }=\sqrt{3}\Rightarrow \frac{h-10}{10\sqrt{3}}=\sqrt{3}$ [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 40 metres.

Alternative Method,

Let BC = h
In $\Delta ACD$

$tan{30}^{\circ }=\frac{10}{x}$

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

$x=10\sqrt{3}=17.32m$ [1 MARK]

In $\Delta ACB$

$tan{60}^{\circ }=\frac{h}{x}$

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

$\sqrt{3}\left(10\sqrt{3}\right)=h$

$h=30m$

Height of hill = h + 10 = 30 + 10

= 40 m [1 MARK]