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

Let's calculate the relative height of the top of the tower with respect to the observer's line of sight:
$$\Delta H=\text{Height of tower}-\text{Height of observer}=4h-h=3h$$

The horizontal distance ($D$) from the observer to the tower is $3\sqrt{3}h\text{meters}$.

Let $\theta$ be the required angle of elevation. Using the tangent trigonometric function:

$$\tan \theta =\frac{\text{Opposite Side}}{\text{Adjacent Side}}=\frac{\Delta H}{D}$$

$$\tan \theta =\frac{3h}{3\sqrt{3}h}=\frac{1}{\sqrt{3}}$$

We know that $\tan 30^{\circ }=\frac{1}{\sqrt{3}}$, hence:

$$\theta =30^{\circ }$$