On the same side of a tower, two objects are located. Observed from the top of the tower, their angles of depression are 45° and 60°. If the height of the tower is 600 m, the distance between the objects is approximately equal to :

Solution(By Examveda Team)

Let DC be the tower and A and B be the objects as shown above.
Given that DC = 600 m, ∠DAC = 45°, ∠DBC = 60°
$$\eqalign{ & \tan {60^ \circ } = \frac{{DC}}{{BC}} \cr & \sqrt 3 = \frac{{600}}{{BC}} \cr & BC = \frac{{600}}{{\sqrt 3 }}\,..........\,\left( {\text{i}} \right) \cr & \tan {45^ \circ } = \frac{{DC}}{{AC}} \cr & 1 = \frac{{600}}{{AC}} \cr & AC = 600\,............\,\left( {{\text{ii}}} \right) \cr} $$
Distance between the objects
$$ = AC = \left( {AC - BC} \right)$$
$$ = 600 - \frac{{600}}{{\sqrt 3 }}$$   [∵ from (1) and (ii)]
$$\eqalign{ & = 600\left( {1 - \frac{1}{{\sqrt 3 }}} \right) \cr & = 600\left( {\frac{{\sqrt 3 - 1}}{{\sqrt 3 }}} \right) \cr & = 600\left( {\frac{{\sqrt 3 - 1}}{{\sqrt 3 }}} \right) \times \frac{{\sqrt 3 }}{{\sqrt 3 }} \cr & = \frac{{600\sqrt 3 \left( {\sqrt 3 - 1} \right)}}{3} \cr & = 200\sqrt 3 \left( {\sqrt 3 - 1} \right) \cr & = 200\left( {3 - \sqrt 3 } \right) \cr & = 200\left( {3 - 1.73} \right) \cr & = 254\,m \cr} $$