A man from the top of a 100 metres high tower observes 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. `100 sqrt(3)`
B. `200// sqrt(3)`
C. `100// sqrt(3)`
D. `200 sqrt(3)`

Answer

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

### Step 1: Understand the Problem
We have a tower of height 100 meters. A man at the top of the tower observes a car moving towards the tower. The angles of depression to the car at two different points are given as 30° and 60°.

### Step 2: Draw the Diagram
1. Draw a vertical line representing the tower (AB) of height 100 meters.
2. Mark point A as the top of the tower and point B as the bottom.
3. Mark the position of the car at two different points, C (when the angle of depression is 30°) and D (when the angle of depression is 60°).
4. Draw horizontal lines from A to C and A to D, creating two right triangles (ABC and ABD).

### Step 3: Apply Trigonometric Ratios
1. For triangle ABC (where the angle of depression is 30°):
- The angle at A is 30°.
- Using the tangent function:
\[ \tan(30°) = \frac{AB}{BC} \]
- Here, \(AB = 100\) meters (height of the tower) and \(BC\) is the distance from the base of the tower to point C (the car's position).
- We know that \(\tan(30°) = \frac{1}{\sqrt{3}}\).
- Therefore:
\[ \frac{1}{\sqrt{3}} = \frac{100}{BC} \]
- Rearranging gives:
\[ BC = 100\sqrt{3} \text{ meters} \]

### Step 4: Calculate the Distance for Triangle ABD
1. For triangle ABD (where the angle of depression is 60°):
- The angle at A is 60°.
- Using the tangent function:
\[ \tan(60°) = \frac{AB}{BD} \]
- Here, \(BD\) is the distance from the base of the tower to point D (the car's new position).
- We know that \(\tan(60°) = \sqrt{3}\).
- Therefore:
\[ \sqrt{3} = \frac{100}{BD} \]
- Rearranging gives:
\[ BD = \frac{100}{\sqrt{3}} \text{ meters} \]

### Step 5: Find the Distance Travelled by the Car
1. The distance travelled by the car (DC) is the difference between the distances BC and BD:
\[ DC = BC - BD \]
\[ DC = 100\sqrt{3} - \frac{100}{\sqrt{3}} \]
2. To simplify, find a common denominator:
\[ DC = 100\sqrt{3} - \frac{100}{\sqrt{3}} = \frac{100 \cdot 3 - 100}{\sqrt{3}} = \frac{200}{\sqrt{3}} \text{ meters} \]

### Final Answer
The distance travelled by the car during this time is:
\[ \frac{200}{\sqrt{3}} \text{ meters} \]