Question

A girl standing on a lighthouse built on a cliff near the seashore, observes two boats due East of the lighthouse. The angles of depression of the two boats are ${30}^{\circ }$ and ${60}^{\circ }$. The distance between the boats is 300 m. Find the distance of the top of the lighthouse from the sea level. (Boats and foot of the lighthouse are in a straight line.)

Answer 1

Let $A$ and $D$ denote the foot of the cliff and the top of the lighthouse respectively.
Let $B$ and $C$ denote the two boats.
Let $h$ metres be the distance of the top of the lighthouse from the sea level.
Let $AB=x$ metres.
Given that $\angle ABD={60}^{\circ },\angle ACD={30}^{\circ }$
In the right angled $△ABD$,

$\tan {60}^{\circ }=\frac{AD}{AB}$

$\Rightarrow AB=\frac{AD}{\tan {60}^{\circ }}$

Thus, $x=\frac{h}{\sqrt{3}}$ ..........(1)

Also, in the right angled $△ACD$, we have

$\tan {30}^{\circ }=\frac{AD}{AC}$

$\Rightarrow AC=\frac{AD}{\tan {30}^{\circ }}\Rightarrow x+300=\frac{h}{\left(\frac{1}{\sqrt{3}}\right)}$

Thus, $x+300=h\sqrt{3}$ ........ (2)

Using (1) in (2), we get:

$\frac{h}{\sqrt{3}}+300=h\sqrt{3}$

$\Rightarrow h\sqrt{3}-\frac{h}{\sqrt{3}}=300$

$\therefore 2h=300\sqrt{3}$.

$\therefore h=150\sqrt{3}m$.

Hence, the height of the lighthouse from the sea level is $150\sqrt{3}m$.