From a point on the ground, the angle of elevation of an aeroplane is initially 45°. After the aeroplane flies horizontally for 15 seconds, the angle of elevation becomes 30°. If the aeroplane is flying at a constant height of 2500 metres above the ground, what is its speed in km/hr?

Introduction / Context:
This is a classic trigonometry and relative motion problem. The aeroplane flies horizontally at a constant height, causing the angle of elevation at a fixed observation point on the ground to change from 45° to 30° over a known time. From this change in viewing geometry, we deduce the horizontal speed of the aeroplane in km/hr.

Given Data / Assumptions:

• Initial angle of elevation of the aeroplane = 45°.
• Final angle of elevation after 15 seconds = 30°.
• Constant height of the aeroplane above the ground = 2500 m.
• The aeroplane moves in a straight horizontal line.
• The observer is fixed at one point on level ground.
• We ignore effects like Earth curvature and assume right-triangle geometry holds.

Concept / Approach:
For a fixed height h and horizontal distance x from the observer, we use the tangent relation:
tan(theta) = h / xAt 45° and 30°, the corresponding distances x1 and x2 can be computed. The change in these distances over 15 seconds gives the horizontal speed. Finally, we convert the speed from m/s to km/hr using the factor 3.6.

Step-by-Step Solution:
Let h = 2500 m.At first, tan(45°) = h / x1 ⇒ 1 = 2500 / x1 ⇒ x1 = 2500 m.After 15 s, tan(30°) = h / x2 ⇒ 1 / √3 = 2500 / x2.So x2 = 2500 * √3 m.Horizontal distance travelled = x2 − x1 = 2500(√3 − 1) m.Time taken = 15 s.Speed in m/s = 2500(√3 − 1) / 15.Multiply by 3.6 to convert to km/hr:Speed = [2500(√3 − 1) / 15] * 3.6 = 600(√3 − 1) km/hr.

Verification / Alternative check:
We can approximate √3 ≈ 1.732. Then:
Speed ≈ 600 * (1.732 − 1) = 600 * 0.732 ≈ 439.2 km/hr.This is a realistic cruising speed for an aeroplane, confirming that the formula and approach are plausible for the scenario.

Why Other Options Are Wrong:
600 km/hr: Ignores the trigonometric adjustment due to the angles and uses an arbitrary value.600(√3 + 1) km/hr and 600 √3 km/hr: These are much larger and do not follow from the distance difference x2 − x1.500(√3 − 1) km/hr: Uses the correct functional form but with an incorrect coefficient due to wrong distance or time handling.

Common Pitfalls:
Students sometimes subtract angles (45° − 30°) and attempt to use a single tangent instead of working with two separate triangles. Another common error is to treat height as changing or to forget to convert from metres per second to kilometres per hour. Careful identification of x1 and x2 and proper time conversion are crucial.

Final Answer:
The speed of the aeroplane is 600(√3 − 1) km/hr.