The angle of elevation of an aeroplane from a point on the ground is 45°. After flying for 15 seconds, the angle of elevation becomes 30°. If the aeroplane is flying at a constant height of 2500 m, what is its speed in km/hr?

Introduction / Context:
This is a height and distance question involving an aeroplane flying at a fixed altitude. The angle of elevation from a point on the ground to the aeroplane changes as the plane moves, and you are given the time taken for this change. You must find the speed of the aeroplane using trigonometric relationships and speed = distance / time, then convert the result into kilometres per hour.

Given Data / Assumptions:

• Constant height of the aeroplane above ground h = 2500 m.
• Initial angle of elevation from the observer = 45 degrees.
• Angle of elevation after 15 seconds = 30 degrees.
• The aeroplane is flying in a horizontal straight line away from the observer.
• Standard values: tan 45 = 1, tan 30 = 1 / sqrt(3).

Concept / Approach:
We model the situation with right angled triangles at two different instants, both having the same vertical height but different horizontal distances from the observer. Using tan θ = height / base, we find the initial and final horizontal distances. Their difference is the horizontal distance travelled during 15 seconds. Dividing by time gives the speed in metres per second, which is then converted to kilometres per hour by multiplying by 3.6.

Step-by-Step Solution:
Let x1 be the initial horizontal distance when the angle is 45 degrees. tan 45 = 2500 / x1 ⇒ 1 = 2500 / x1 ⇒ x1 = 2500 m. Let x2 be the horizontal distance after 15 seconds when the angle is 30 degrees. tan 30 = 2500 / x2 ⇒ 1 / sqrt(3) = 2500 / x2 ⇒ x2 = 2500√3 m. Horizontal distance travelled in 15 seconds = x2 − x1 = 2500√3 − 2500 = 2500(√3 − 1) m. Speed in m/s = distance / time = 2500(√3 − 1) / 15. Convert to km/hr by multiplying by 3.6: speed = [2500(√3 − 1) / 15] × 3.6 = 600(√3 − 1) km/hr.

Verification / Alternative check:
Using √3 ≈ 1.732, we get (√3 − 1) ≈ 0.732. Then 600 × 0.732 ≈ 439.2 km/hr, which is a realistic cruising speed for a smaller aircraft. Reversing the process, 439.2 km/hr divided by 3.6 is about 122 m/s. Over 15 seconds, the plane travels roughly 1830 m, which is close to 2500(√3 − 1) ≈ 1830 m, confirming the consistency of the calculation.

Why Other Options Are Wrong:

• 600 km/hr: This assumes a much smaller horizontal distance change and ignores the factor (√3 − 1).
• 600(√3 + 1) km/hr and 600√3 km/hr: These values are significantly larger and are not supported by the geometry of the two triangles.
• 500√3 km/hr: This uses an incorrect combination of the height and trig values and does not match the distance travelled in 15 seconds.

Common Pitfalls:
Common mistakes include swapping the roles of tan 30 and tan 45, or incorrectly assuming the aeroplane is directly overhead at one of the instants. Some learners forget to subtract distances in the correct order to get the distance travelled. Others make unit conversion errors when converting from m/s to km/hr. Always handle each step carefully and keep track of your units.

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