A helicopter, at an altitude of 1500 metre, finds that two ships are sailing towards it, in the same direction. The angles of depression of the ships as observed from the helicopter are `60^@ and 30^@` respectively. Distance between the two ships, in metre is

Options

A. `1000sqrt3`
B. `(1000)/(sqrt3)`
C. `500sqrt3`
D. `(500)/(sqrt3)`

Answer

To solve the problem, we need to find the distance between two ships observed from a helicopter at an altitude of 1500 meters, given the angles of depression to the ships are 60° and 30° respectively.

### Step-by-Step Solution:

1. **Understand the Geometry**:
- Let the helicopter be at point A, the first ship at point D, and the second ship at point E.
- The altitude of the helicopter (AC) is 1500 meters.
- The angles of depression to the ships are 60° (for ship D) and 30° (for ship E).

2. **Draw the Right Triangles**:
- From point A (helicopter), draw a vertical line down to point C (the ground directly below A).
- Draw horizontal lines from C to D and C to E, forming right triangles ACD and ACE.

3. **Using Trigonometry for Ship D**:
- In triangle ACD, where angle CAD = 60°:
\[ \tan(60°) = \frac{AC}{CD} \]
\[ \sqrt{3} = \frac{1500}{CD} \]
\[ CD = \frac{1500}{\sqrt{3}} = 500\sqrt{3} \text{ meters} \]

4. **Using Trigonometry for Ship E**:
- In triangle ACE, where angle CAE = 30°:
\[ \tan(30°) = \frac{AC}{CE} \]
\[ \frac{1}{\sqrt{3}} = \frac{1500}{CE} \]
\[ CE = 1500\sqrt{3} \text{ meters} \]

5. **Finding the Distance Between the Two Ships**:
- The distance DE between the two ships can be found by subtracting the distances from C to D and C to E:
\[ DE = CE - CD \]
\[ DE = 1500\sqrt{3} - 500\sqrt{3} = (1500 - 500)\sqrt{3} = 1000\sqrt{3} \text{ meters} \]

### Final Answer:
The distance between the two ships is \(1000\sqrt{3}\) meters.