To solve this problem, we need to find the horizontal distance between two points on the ground from which the angles of depression \(45^\circ\) and \(30^\circ\) are observed from the top of a 50 m tall building.
We will use trigonometry to solve this. Let's denote:
For a \(45^\circ\) angle of depression, we use the tangent function:
\(\tan(45^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{H}{x}\)
Since \(\tan(45^\circ) = 1\), we have:
\(1 = \frac{50}{x} \Rightarrow x = 50 \text{ m}\)
For a \(30^\circ\) angle of depression, we use the tangent function:
\(\tan(30^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{H}{y}\)
Since \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\), we have:
\(\frac{1}{\sqrt{3}} = \frac{50}{y} \Rightarrow y = 50\sqrt{3} \text{ m}\)
The distance \(BC\) is the difference between \(AC\) and \(AB\):
\(BC = y - x = 50\sqrt{3} - 50\)
Factoring out 50 gives:
\(BC = 50(\sqrt{3} - 1) \text{ m}\)
Thus, the distance between the two points is \(\boldsymbol{50(\sqrt{3} - 1) \, \text{m}}\).