A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower at a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point. 

To solve this problem, we will use trigonometric relationships and analyze the change in the car's position with respect to the tower.

### Step-by-step Solution:

1. **Understanding the Problem:**

   - Let the height of the tower be \( h \).
   - Let the initial distance of the car from the foot of the tower when the angle of depression is \( 30^\circ \) be \( x_1 \).
   - After 6 seconds, the car moves closer, and the angle of depression changes to \( 60^\circ \). Let the new distance from the foot of the tower be \( x_2 \).

2. **Trigonometric Relations:**

   Since the angle of depression is measured from the horizontal line (parallel to the ground) at the top of the tower, the corresponding angles of elevation of the car from the foot of the tower are also \( 30^\circ \) and \( 60^\circ \).

   Using the tangent function for both angles:

   \[
   \tan(30^\circ) = \frac{h}{x_1}
   \]

   \[
   \tan(60^\circ) = \frac{h}{x_2}
   \]

   We know that:
   \[
   \tan(30^\circ) = \frac{1}{\sqrt{3}}, \quad \tan(60^\circ) = \sqrt{3}
   \]

   Thus:

   \[
   \frac{1}{\sqrt{3}} = \frac{h}{x_1} \implies x_1 = h \sqrt{3}
   \]

   \[
   \sqrt{3} = \frac{h}{x_2} \implies x_2 = \frac{h}{\sqrt{3}}
   \]

3. **Finding the Distance Covered in 6 Seconds:**

   The car travels from \( x_1 \) to \( x_2 \) in 6 seconds. The distance covered by the car in 6 seconds is:

   \[
   x_1 - x_2 = h \sqrt{3} - \frac{h}{\sqrt{3}}
   \]

   Let's simplify this expression:

   \[
   x_1 - x_2 = h \left( \sqrt{3} - \frac{1}{\sqrt{3}} \right)
   \]

   \[
   x_1 - x_2 = h \left( \frac{3}{\sqrt{3}} - \frac{1}{\sqrt{3}} \right)
   \]

   \[
   x_1 - x_2 = h \left( \frac{2}{\sqrt{3}} \right)
   \]

   \[
   x_1 - x_2 = \frac{2h}{\sqrt{3}}
   \]

4. **Calculate the Speed of the Car:**

   The speed \( v \) of the car is given by the distance covered divided by the time taken:

   \[
   v = \frac{x_1 - x_2}{6} = \frac{\frac{2h}{\sqrt{3}}}{6} = \frac{2h}{6\sqrt{3}} = \frac{h}{3\sqrt{3}}
   \]

5. **Finding the Time to Reach the Foot of the Tower:**

   The remaining distance the car needs to travel to reach the foot of the tower from \( x_2 \) is \( x_2 = \frac{h}{\sqrt{3}} \).

   Time \( t \) taken to cover this distance at speed \( v \) is:

   \[
   t = \frac{x_2}{v} = \frac{\frac{h}{\sqrt{3}}}{\frac{h}{3\sqrt{3}}}
   \]

   Simplify the expression:

   \[
   t = \frac{h}{\sqrt{3}} \times \frac{3\sqrt{3}}{h}
   \]

   \[
   t = 3
   \]

### Conclusion:
The time taken by the car to reach the foot of the tower from the point where the angle of depression is \(60^\circ\) is **3 seconds**.

---

Would you like a detailed explanation of any step? Or do you have any other questions? Here are some related questions:

1. How does the concept of angle of depression apply to real-life situations?
2. What is the difference between angle of elevation and angle of depression?
3. Can you explain the trigonometric functions used in the solution?
4. How can the speed of the car be calculated differently?
5. What are the practical applications of trigonometry in navigation?

**Tip:** Remember to carefully consider the angle of depression or elevation when solving trigonometric problems involving heights and distances.

Learn how to calculate the time taken by a car to reach the foot of a tower using trigonometric angles of depression. Detailed step-by-step solution with trigonometric formulas and practical application of angles of depression. Suitable for high school students.

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