Question

The angles of depression of two ships from the top of a lighthouse and on the same side of it are found to be ${45}^{\circ }$ and ${30}^{\circ }$ respectively. If the ships are $200m$ apart, find the height of the lighthouse.

Answer 1

Let $CD$ be the lighthouse and $A$ and $B$ be the positions of the two ships.
$AB=200m$ (Given)

Suppose $CD=hm$ and $BC=xm$
Now, $\angle DAC=\angle ADE={30}^{\circ }$ (Alternate interior angles)

$\angle DBC=\angle EDB={45}^{\circ }$ (Alternate interior angles)

In right $△BCD,$ we have

$\tan {45}^{\circ }=\frac{CD}{BC}$

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

$\Rightarrow h=x$

In right $△ACD,$ we have

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

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+200}$

$\Rightarrow h\sqrt{3}=x+200$

$\Rightarrow h\sqrt{3}=h+200$

$\Rightarrow h=\frac{200}{\sqrt{3}-1}$

$\therefore h=100(\sqrt{3}+1)m$