Let us learn real-life situations based on the angle of elevation.
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 the point \(A\) to \(C\).
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 points \(A\) and \(C\) is \(500\sqrt{3} \ m\).