The height of a tower is $4h\text{meter}$ and the height of the observer is $h\text{meter}$. He finds the top of the tower from a distance $\sqrt{3}h\text{meter}$ from the tower. Find the angle of elevation of the tower as seen by the observer.

Let the angle of elevation seen from the horizontal line of sight of the observer be $\theta$.
* The effective perpendicular height above the observer's eye-level line is:

$$\text{Height}=4h-h=3h\text{meters}$$

* The horizontal base distance remains:

$$\text{Base}=\sqrt{3}h\text{meters}$$

Using the trigonometric ratio for tangent:

$$\tan \theta =\frac{\text{Perpendicular}}{\text{Base}}=\frac{3h}{\sqrt{3}h}=\frac{3}{\sqrt{3}}=\sqrt{3}$$

Since $\tan (60^{\circ })=\sqrt{3}$, the angle of elevation $\theta$ is $60^{\circ }$.