# The angle of depression of two ships from the tower are 30Â° and 45Â° towards east if the ships are apart from each other by hundred metre then the height of the tower isâ€‹

Let AB be the light house C and D are two ships whose angles of depression on A are 30° and 45° respectively
$\\ here\ \angle A C B=\angle X A C=30^{\circ}, \angle A D B=\angle Y A D=45^{\circ} and \mathrm{CD}=100 \mathrm{m}\\ Let\ A B=h \ and\ C B=x then B C=(100-x) m \\ Now \ in \ \Delta A C B, \tan \theta=\frac{A B}{C B} \Rightarrow \tan 30^{\circ}=\frac{h}{x} \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x} \Rightarrow x=\sqrt{3} h \ldots \ldots (i)\\ Similarly\ in\ \Delta A D B\\ \tan 45^{\circ}=\frac{A B}{B D} \Rightarrow 1=\frac{h}{100-x} \Rightarrow x=100-h \ldots \ldots(ii)$

$\\ From \ (i)\ and\ (ii)\\ \sqrt{3} h=100-h \Rightarrow(\sqrt{3}-1) h=100$

$h=\frac{100}{\sqrt{3}+1} \\ \Rightarrow h=\frac{100(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)} = \frac{100(\sqrt{3}-1)}{3-1} = \frac{100(\sqrt{3}-1)}{2} \\\Rightarrow h = 50(\sqrt{3}- \therefore \text{height of light house} =50(\sqrt{3}-1) m$

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