A man on the top of a vertical tower observes a car moving at a uniform speed towards the tower on a horizontal road. If it takes 18 min for the angle of depression of the car to change from 30$$^\circ$$ to 45$$^\circ$$, then the time taken (in min) by the car to reach the foot of the tower is:

Let us assume that the tower is perfectly vertical and its height is $$h$$ metres. The car is moving on a horizontal road which we shall regard as level ground. From the top of the tower a man looks down at the car; therefore the angle of depression seen by the man is numerically equal to the angle of elevation that the car would see if it looked up at the top of the tower (alternate angles made by a transversal with two parallel horizontal lines).

At an initial instant the angle of depression is $$30^\circ$$. Let the horizontal distance of the car from the foot of the tower at that moment be $$x_1$$ metres. By simple right-triangle trigonometry in the vertical plane that contains the tower and the car we have the tangent relation:

$$\tan 30^\circ \;=\;\frac{\text{opposite side}}{\text{adjacent side}}\;=\;\frac{h}{x_1}.$$

The numerical value of $$\tan 30^\circ$$ is $$\frac{1}{\sqrt3}$$, so

$$\frac{1}{\sqrt3} \;=\;\frac{h}{x_1}.$$

Cross-multiplying gives

$$x_1 \;=\;\sqrt3\,h.$$

After the car has moved nearer, the angle of depression becomes $$45^\circ$$. Let the new horizontal distance be $$x_2$$ metres. Using the same tangent formula, now with $$45^\circ$$, we write

$$\tan 45^\circ \;=\;\frac{h}{x_2}.$$

Since $$\tan 45^\circ = 1$$, we have immediately

$$1 \;=\;\frac{h}{x_2}\quad\Longrightarrow\quad x_2 \;=\;h.$$

Thus the car travels the horizontal distance

$$x_1 - x_2 \;=\;\sqrt3\,h - h \;=\;h(\sqrt3 - 1)$$

while the angle of depression changes from $$30^\circ$$ to $$45^\circ$$. According to the statement of the problem this travel takes $$18$$ minutes. If the uniform speed of the car be $$v$$ metres per minute, then

$$v \;=\;\frac{\text{distance}}{\text{time}}\;=\;\frac{h(\sqrt3 - 1)}{18}.$$

After this instant the car still has to cover the remaining horizontal distance $$x_2$$ (which equals $$h$$) to reach the foot of the tower. The time $$t$$ required for that final leg is found from

$$t \;=\;\frac{\text{distance still to go}}{\text{same uniform speed}} \;=\;\frac{h}{v}.$$

Substituting the expression for $$v$$ just obtained, we get

$$t \;=\;\frac{h}{\dfrac{h(\sqrt3 - 1)}{18}}.$$

We may cancel the common factor $$h$$ in numerator and denominator, giving

$$t \;=\;18\,\frac{1}{\sqrt3 - 1}.$$

To rationalise the denominator we multiply the numerator and the denominator by the conjugate $$(\sqrt3 + 1):$$

$$t \;=\;18\,\frac{\sqrt3 + 1}{(\sqrt3 - 1)(\sqrt3 + 1)}.$$

The product in the denominator is a difference of squares:

$$(\sqrt3 - 1)(\sqrt3 + 1) \;=\;(\sqrt3)^2 - 1^2 \;=\;3 - 1 \;=\;2.$$

Therefore

$$t \;=\;18\,\frac{\sqrt3 + 1}{2} \;=\;9(\sqrt3 + 1).$$

Since all quantities were in minutes, the required time for the car to reach the foot of the tower is $$9(1 + \sqrt3)$$ minutes.

Hence, the correct answer is Option D.