Question

abhimanyu observes two ships from aeroplane at an altitude of 5000m sailing in the same direction the angles of depression of the ships as observed are 45 and 30 find the distance between the ships​

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.

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

$In \:\triangle ABC , \:we \:have$

$tan 45\degree = \frac{AB}{CB}$

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

$In \:\triangle ABD, \:we \:have$

$tan 30\degree = \frac{AB}{DB}$

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

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

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

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

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

$\implies DC = 5000\times 0.732$

$\implies DC = 3660 \:m$

Therefore.,

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

•••♪