Question

A Abhimanyu observes two ships from an aeroplane at an altitude
of 5000 m sailing in the same direction. The angles of depression
of the ships as observed are 45° and 30°. Find the distance
between ship​

Answer 1

Let D and C be the positions of two ships .

Distance between two ships = CD ,

Let the observer Abhimanyu be at A ,in Aeroplane.

It is given that AB = 5000 m and the angles of depression from A of D and C are 30° and 45° respectively .

$\angle {ADB} = 30\degree \:and \:\angle {ACB} = 45\degree$

$In \: \triangle ABC, \:we \:have \\tan \:45\degree = \frac{AB}{CB}$

$\implies 1 = \frac{5000}{CB}$

$\implies CB = 5000 \:m$

$In \: \triangle ABD ,we \:have \\tan \:30\degree = \frac{AB}{DB}$

$\implies \frac{1}{\sqrt{3}} = \frac{5000}{DC + CB}$

$\implies DC + CB = 5000\sqrt{3}$

$\implies DC + 5000 = 5000\sqrt{3}$

$\implies DC = - 5000 +5000\sqrt{3}$

$\implies DC = (\sqrt{3} - 1)5000$

$\implies DC = (1.732 - 1)5000$

$\implies DC = 0.732 \times 5000$

$\implies DC = 0.732 \times 5000$

$\implies DC = 3660 \:m$

Therefore.,

$\red { Distance \: between \:two \:ships } \green {= 3660 \:m}$

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