A highway leads to the foot of 300 m high tower. An observatory is set at the top of the tower. It sees a car moving towards it at an angle of depression of `30^(@)`. After 15 seconds, angle of depression becomes `60^(@)`. Find the distance travelled by the car during this time.

Answer

To solve the problem, we need to find the distance traveled by the car during the time it takes for the angle of depression to change from 30 degrees to 60 degrees. We will use trigonometry to find the distances involved.

### Step-by-Step Solution:

1. **Identify the height of the tower:**
The height of the tower (AB) is given as 300 m.

2. **Set up the angles of depression:**
- When the angle of depression is 30 degrees, we will denote the distance from the base of the tower to the car as \( DB \).
- When the angle of depression is 60 degrees, we will denote the distance from the base of the tower to the car as \( CB \).

3. **Use trigonometric ratios:**
- For the angle of depression of 30 degrees:
\[ \tan(30^\circ) = \frac{AB}{DB} \]
\[ \tan(30^\circ) = \frac{300}{DB} \implies DB = \frac{300}{\tan(30^\circ)} = \frac{300}{\frac{1}{\sqrt{3}}} = 300\sqrt{3} \text{ m} \]

- For the angle of depression of 60 degrees:
\[ \tan(60^\circ) = \frac{AB}{CB} \]
\[ \tan(60^\circ) = \frac{300}{CB} \implies CB = \frac{300}{\tan(60^\circ)} = \frac{300}{\sqrt{3}} = 100\sqrt{3} \text{ m} \]

4. **Calculate the distance traveled by the car:**
The distance traveled by the car (DC) during the 15 seconds is the difference between the two distances calculated:
\[ DC = DB - CB = 300\sqrt{3} - 100\sqrt{3} = 200\sqrt{3} \text{ m} \]

5. **Final answer:**
The distance traveled by the car during the time is \( 200\sqrt{3} \) meters.