An aeroplane at an altitude 1500m, find that the two ships are sailing towards it in the same direction. The angle of depression as observed from the airplane is 45∘ and 30∘ respectively. Find the distance (in m) between two ships.

To solve this problem, we can use trigonometry. Let's break it down step by step:

1. **Define the situation:**
- An airplane is at an altitude of 1500 meters.
- The angle of depression to Ship 1 is 45°.
- The angle of depression to Ship 2 is 30°.
- The ships are sailing towards the airplane in the same direction.

2. **Label the variables:**
- Let **A** be the point where the airplane is located.
- Let **B** and **C** be the points on the ground directly below Ship 1 and Ship 2, respectively.
- The distance from the airplane to Ship 1 is the horizontal distance from **A** to **B**.
- The distance from the airplane to Ship 2 is the horizontal distance from **A** to **C**.

3. **Use of tangent formula:**
The formula for tangent in a right triangle is:
\[
\tan(\theta) = \frac{\text{Opposite side}}{\text{Adjacent side}}
\]
Where the "opposite side" is the height of the airplane (1500 meters), and the "adjacent side" is the horizontal distance.

4. **For Ship 1:**
The angle of depression is 45°, so:
\[
\tan(45^\circ) = \frac{1500}{x_1}
\]
Since \(\tan(45^\circ) = 1\), we have:
\[
1 = \frac{1500}{x_1}
\]
Therefore, the horizontal distance to Ship 1 is:
\[
x_1 = 1500 \, \text{m}
\]

5. **For Ship 2:**
The angle of depression is 30°, so:
\[
\tan(30^\circ) = \frac{1500}{x_2}
\]
We know that \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\), so:
\[
\frac{1}{\sqrt{3}} = \frac{1500}{x_2}
\]
Therefore:
\[
x_2 = 1500 \cdot \sqrt{3} \approx 1500 \cdot 1.732 = 2598 \, \text{m}
\]

6. **Find the distance between the two ships:**
The two ships are sailing towards each other, so the distance between them is the difference in the horizontal distances:
\[
\text{Distance between ships} = x_2 - x_1 = 2598 - 1500 = 1098 \, \text{m}
\]

Thus, the distance between the two ships is **1098 meters**.