Step 1: Recognise the special right-triangle ratio.
Rather than computing $\tan\theta$ and then looking up the angle, notice the given numbers directly match the classic $30^{\circ}\text{-}60^{\circ}\text{-}90^{\circ}$ triangle, whose sides are always in the fixed ratio $1 : \sqrt{3} : 2$ opposite the angles $30^{\circ}, 60^{\circ}, 90^{\circ}$.
Step 2: Match the given sides to this ratio.
The base (distance of car from tower) is $10\sqrt{3}\text{ m}$ and the height (perpendicular) is $30\text{ m}$. Write both in terms of a common scale factor $k$:
\[ 10\sqrt{3} = k\sqrt{3} \implies k = 10 \]
\[ 30 = 3k = 3(10) = 30 \]
Both sides scale consistently with $k = 10$ against the ratio parts $\sqrt{3}$ and $3$ (which is $\sqrt{3} \times \sqrt{3}$, corresponding to a triangle where the side opposite $60^{\circ}$ is $\sqrt{3}$ times the side opposite $30^{\circ}$).
Step 3: Identify which angle is opposite the height.
In this right triangle, the height $30\text{ m}$ (the longer leg) is opposite the angle of elevation $\theta$, and the base $10\sqrt{3}\text{ m}$ (the shorter leg) is opposite the remaining acute angle. Since the longer leg is $\sqrt{3}$ times the shorter leg here
\[ \frac{30}{10\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3} \]
this matches the leg-ratio pattern of a $30^{\circ}\text{-}60^{\circ}\text{-}90^{\circ}$ triangle where the angle opposite the longer leg is $60^{\circ}$.
Step 4: State the angle.
So the angle of elevation $\theta = 60^{\circ}$.
Final Answer:
The angle of elevation of the top of the tower from the car is $60^{\circ}$, which corresponds to option (D).
\[ \boxed{60^{\circ}} \]