Two ships are approaching a lighthouse from opposite directions, The angles of depression of the two ships from the top of a lighthouse are $ 30^{\circ} $ and $ 45^{\circ} $. If the distance between the two ships is 100 metres, find the height of the lighthouse. (Use $ \sqrt{3}=1.732 $ )

Given:

Two ships are approaching a lighthouse from opposite directions, The angles of depression of the two ships from the top of a lighthouse are \( 30^{\circ} \) and \( 45^{\circ} \). The distance between the two ships is 100 metres.

To do:

We have to find the height of the lighthouse.

Solution:

Let BC be x and the height of the lighthouse be h.

This implies,

$BD=100-x\ m$

$\angle C=30^o$ and $\angle D=45^o$

In triangle ABC,

$tan 30^o=\frac{x}{h}$

$\frac{1}{\sqrt3}=\frac{h}{x}$

$x=\sqrt{3}h$   (on cross multiplication)----(i)

$tan 45^o=\frac{h}{100-x}$

$1=\frac{h}{100-x}$

$h=100-x$

$x=100-h$   (on cross multiplication)----(ii)

Substitute $x=100-h$ in equation (i)

$100-h=\sqrt{3}h$

$h+\sqrt{3}h=100$

$h(1+1.732)=100$

$h=\frac{100}{2.732}$

$h=36.6$

The height of the lighthouse is 36.6 m.

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Updated on: 10-Oct-2022

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