A helicopter flying at an altitude of 1500 m observes two ships sailing towards it in the same straight line. The angles of depression of the ships are 60° and 30° respectively. What is the distance between the two ships in metres?

Introduction / Context:
Here a helicopter hovers at a fixed altitude and observes two ships that are on the same straight line, sailing towards the point directly beneath the helicopter. The angles of depression to the two ships are 60 degrees and 30 degrees. Using these angles and the fixed altitude, we can find each ship's horizontal distance from the helicopter's projection on the water surface and then compute how far apart the ships are from each other.

Given Data / Assumptions:

• Altitude of the helicopter above sea level = 1500 m.
• Angles of depression to the two ships = 60 degrees and 30 degrees.
• The ships are on the same straight line and on the same side relative to the helicopter's vertical projection.
• Angles of depression equal corresponding angles of elevation from the ships.
• Standard values: tan 60 = sqrt(3), tan 30 = 1 / sqrt(3).

Concept / Approach:
Let the point directly below the helicopter on the water surface be P. Let the nearer ship be at distance x1 from P with angle of elevation 60 degrees and the farther ship at distance x2 from P with angle of elevation 30 degrees. For each triangle, tan θ = height / base. We compute x1 and x2 from the known height 1500 m. The distance between the ships is the absolute difference |x2 − x1| because they lie along the same horizontal line on the same side.

Step-by-Step Solution:
Let x1 be the horizontal distance to the ship seen at angle of depression 60 degrees. tan 60 = 1500 / x1 ⇒ sqrt(3) = 1500 / x1 ⇒ x1 = 1500 / sqrt(3) m. Let x2 be the horizontal distance to the ship seen at angle of depression 30 degrees. tan 30 = 1500 / x2 ⇒ 1 / sqrt(3) = 1500 / x2 ⇒ x2 = 1500√3 m. Distance between the two ships = x2 − x1 (since x2 > x1). So distance = 1500√3 − 1500 / √3. Factor: 1500(√3 − 1 / √3) = 1500((3 − 1) / √3) = 1500 × 2 / √3 = 3000 / √3. Note that 3000 / √3 = 1000 × (3 / √3) = 1000√3 m.

Verification / Alternative check:
Using √3 ≈ 1.732, x1 ≈ 1500 / 1.732 ≈ 866 m, and x2 ≈ 1500 × 1.732 ≈ 2598 m. Their difference is about 1732 m, which is close to 1000√3 ≈ 1732 m. Both triangles are consistent with tan 60 and tan 30 using the same height of 1500 m, which confirms our calculations.

Why Other Options Are Wrong:

• 1000/√3 m and 500/√3 m: These values are much smaller than the actual separation determined by the geometry.
• 500√3 m: This is roughly half the correct separation and does not correspond to the difference between 1500√3 and 1500 / √3.
• 1500√3 m: This is just the distance from P to the farther ship, not the distance between the two ships.

Common Pitfalls:
Mistakes often arise from confusing which ship corresponds to which angle. The nearer ship must have the larger angle (60 degrees) and thus the smaller base distance. Another trap is to add the two distances instead of subtracting them, which would imply the ships are on opposite sides of the helicopter, contradicting the statement that they sail towards it along the same direction. Careful interpretation of the problem text is essential.

Final Answer:
The distance between the two ships is 1000√3 m.