Theory:

Let us learn real-life situations based on the angle of elevation.
Example:
The altitude of the plane from point \(C\) on the ground is \(1500 \ m\). If the angle of elevation of the person from point \(A\) is \(60^{\circ}\), then find the distance of the person from point \(A\) to \(C\).
 
Solution:
 
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Let \(A\) be the position of the person, \(B\) be the position of the airplane, and \(C\) be the point.
 
Then, from the given data, we have:
 
\(\theta = 60^{\circ}\) and \(BC = 1500 \ m\)
 
In the right triangle \(ACB\), we have:
 
\(tan \ \theta = \frac{BC}{AC}\)
 
\(tan \ 60^{\circ} = \frac{1500}{AC}\)
 
\(AC = \frac{1500}{tan \ 60^{\circ}}\)
 
\(AC = \frac{1500}{\sqrt{3}}\)
 
\(AC = 500\sqrt{3}\)
 
Therefore, the distance between the points \(A\) and \(C\) is \(500\sqrt{3} \ m\).