### 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 the point $$A$$ to $$C$$.

Solution:

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 $$500\sqrt{3} \ m$$.