An aeroplane flies horizontally at a constant height of 3 km above the ground. From a point on the ground the angle of elevation of the plane is initially 60°. After 15 seconds of flight, the angle of elevation becomes 30°. Taking √3 = 1.732, what is the speed of the aeroplane in m/s?

Introduction / Context:
This is a standard height and distance problem using trigonometry. The horizontal speed of the aeroplane is determined from the change in angle of elevation observed from a fixed point on the ground. The key idea is to convert angle information into horizontal distances using tangent and then relate distance change to time to find speed.

Given Data / Assumptions:

• Height of plane above ground: 3 km.
• Initial angle of elevation: 60°.
• Angle of elevation after 15 seconds: 30°.
• Plane flies horizontally at constant height.
• Approximation: √3 = 1.732.
• Need speed in metres per second.

Concept / Approach:
At both observation times, the plane, the ground point and the vertical from the plane form right triangles. Using tan θ = opposite / adjacent, we can find the horizontal distances from the point to the plane at each instant. The difference of these distances over the time interval gives the horizontal speed. Finally, convert from kilometres to metres.

Step-by-Step Solution:
Let h = 3 km be the constant height of the plane. Initially, angle of elevation θ₁ = 60°. Let the horizontal distance be d₁. Then tan 60° = h / d₁ ⇒ √3 = 3 / d₁ ⇒ d₁ = 3 / √3 km = √3 km. After 15 seconds, angle θ₂ = 30°. Let the new distance be d₂. tan 30° = h / d₂ ⇒ 1/√3 = 3 / d₂ ⇒ d₂ = 3√3 km. Horizontal distance travelled in 15 s is Δd = d₂ − d₁ = 3√3 − √3 = 2√3 km. Using √3 ≈ 1.732, Δd ≈ 2 × 1.732 km = 3.464 km. Convert to metres: 3.464 km = 3464 m. Speed v = distance / time = 3464 m / 15 s ≈ 230.933 m/s. Rounded to two decimal places, v ≈ 230.93 m/sec.
Verification / Alternative check:
You can recompute using more accurate values of √3 and verify that the speed remains extremely close to 230.93 m/s, which confirms that rounding √3 to 1.732 is acceptable for this problem.

Why Other Options Are Wrong:
230.63 m/s and 235.85 m/s arise from slight miscalculations of distances or using approximate values incorrectly. 236.25 m/s assumes a different height-to-distance relationship, and 210.50 m/s is too low given the large change in horizontal position in just 15 seconds.

Common Pitfalls:
Typical mistakes include using km/h instead of m/s, forgetting to convert kilometres to metres, or mixing up which distance corresponds to which angle. Remember that smaller angle of elevation corresponds to a larger horizontal distance because the plane has moved farther away along the ground.

Final Answer:
The speed of the aeroplane is approximately 230.93 m/sec.