The angles of depression of two ships from the top of a light house and on the same side

Solution:

Let CD be the the light house and A and B be the positions of the two ships.

AB = 200 m    (Given)

Suppose CD = h m and BC = x m

Now,

∠">∠∠DAC = ∠">∠∠ADE = 30º        (Alternate angles)

∠">∠∠DBC = ∠">∠∠EDB = 45º        (Alternate angles)

In right ∆BCD,

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

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

$\Rightarrow x=h \quad \ldots$. (1)

In right ∆ACD,

$\tan 30^{\circ}=\frac{\mathrm{CD}}{\mathrm{AC}}$

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

$\Rightarrow \sqrt{3} h=x+200 \quad \ldots(2)$

From (1) and (2), we get

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

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

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

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

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

$\Rightarrow h=\frac{200(\sqrt{3}+1)}{2}=100(\sqrt{3}+1) \mathrm{m}$

Hence, the height of the light house is $100(\sqrt{3}+1) \mathrm{m}$.