A Abhimanyu observes two ships from an aeroplane at an altitudeof 5000

1.	A Abhimanyu observes two ships from an aeroplane at an altitudeof 5000 m sailing in the same direction. The angles of depressionof the ships as observed are 45° and 30°. Find the distancebetween ship
Answer» 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., $\red { <klux>DISTANCE</klux> \: between \:two \:ships } \green {= 3660 \:m}$ •••♪

A Abhimanyu observes two ships from an aeroplane at an altitudeof 5000 m sailing in the same direction. The angles of depressionof the ships as observed are 45° and 30°. Find the distancebetween ship

Answer»

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 .