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 :

Answer: Option D

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