A man from the top of a 100 metres high tower sees a car moving towards the tower at an angle of depression of `30^@` . After some time, the angle of depression becomes `60^@` .The distance (in metres) travelled by the car during this time is

Options

A. `100sqrt3`
B. `(200sqrt3)/3`
C. `(100sqrt3)/3`
D. `200sqrt3`

Explanation

To solve the problem step by step, we will use trigonometric ratios and properties of right triangles.

### Step 1: Understand the Problem
We have a tower of height 100 meters. A man at the top of the tower sees a car moving towards the tower. The angle of depression to the car changes from 30 degrees to 60 degrees.

### Step 2: Draw the Diagram
Draw a vertical line representing the tower (AB) with point A at the top and point B at the base. From point A, draw two lines representing the line of sight to the car at two different angles of depression (30 degrees and 60 degrees). Let the positions of the car be C and D for the two angles.

### Step 3: Identify the Right Triangles
From point A (top of the tower), draw horizontal line AC and AD to represent the line of sight to the car at angles of depression of 30 degrees and 60 degrees respectively. The height of the tower (AB) is 100 meters.

### Step 4: Use Trigonometric Ratios
For angle of depression of 30 degrees:
- In triangle ABC, we have:
\[ \tan(30^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BC} \]
\[ \tan(30^\circ) = \frac{100}{x + y} \quad \text{(where \(x + y\) is the distance from the tower to the car at 30 degrees)} \]
Since \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\):
\[ \frac{1}{\sqrt{3}} = \frac{100}{x + y} \]
Rearranging gives:
\[ x + y = 100\sqrt{3} \quad \text{(Equation 1)} \]

For angle of depression of 60 degrees:
- In triangle ABD, we have:
\[ \tan(60^\circ) = \frac{AB}{BD} = \frac{100}{y} \]
Since \(\tan(60^\circ) = \sqrt{3}\):
\[ \sqrt{3} = \frac{100}{y} \]
Rearranging gives:
\[ y = \frac{100}{\sqrt{3}} \quad \text{(Equation 2)} \]

### Step 5: Substitute Equation 2 into Equation 1
Substituting the value of \(y\) from Equation 2 into Equation 1:
\[ x + \frac{100}{\sqrt{3}} = 100\sqrt{3} \]
Rearranging gives:
\[ x = 100\sqrt{3} - \frac{100}{\sqrt{3}} \]

### Step 6: Simplify the Expression for x
To simplify \(x\):
\[ x = 100\left(\sqrt{3} - \frac{1}{\sqrt{3}}\right) \]
Finding a common denominator:
\[ x = 100\left(\frac{3 - 1}{\sqrt{3}}\right) = 100\left(\frac{2}{\sqrt{3}}\right) \]
Thus:
\[ x = \frac{200}{\sqrt{3}} \]

### Step 7: Find the Distance Travelled by the Car
The total distance travelled by the car from position C to position D is equal to \(y\):
\[ \text{Distance travelled} = y = \frac{100}{\sqrt{3}} \]

### Final Calculation
To express the distance in a more usable form:
\[ y = \frac{100\sqrt{3}}{3} \text{ meters} \]

### Conclusion
The distance travelled by the car during this time is:
\[ \text{Distance} = \frac{200}{\sqrt{3}} \approx 115.47 \text{ meters} \]