An aeroplane when flying at a height of 3125 m from the ground passes vertically below another plane at an instant when the angles of elevation of the two planes from the same point on the ground are 30° and 60° respectively. Find the distance between the two planes at that instant.
Generated by claude-sonnet-4-6 · 2026-06-15 10:37 · grounding rag
Model Answer
Let the point on the ground be O, and let the horizontal distance from O to the point directly below both planes be $d$ metres.
For the lower plane (height = 3125 m, angle of elevation = 30°):
$$\tan 30° = \frac{3125}{d} \implies \frac{1}{\sqrt{3}} = \frac{3125}{d} \implies d = 3125\sqrt{3} \text{ m}$$
For the upper plane (height = $h$ m, angle of elevation = 60°):
$$\tan 60° = \frac{h}{d} \implies \sqrt{3} = \frac{h}{3125\sqrt{3}} \implies h = 3125\sqrt{3} \times \sqrt{3} = 9375 \text{ m}$$
Distance between the two planes:
$$= 9375 - 3125 = \boxed{6250 \text{ m}}$$
Source: Chapter 9, Heights and Distances
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Explanation
- Both planes fly vertically above the same point, so both share the same horizontal distance $d$ from the observer.
- Use $\tan\theta = \frac{\text{height}}{d}$ for each plane separately.
- Find $d$ from the lower plane's data, then find the upper plane's height using the same $d$.
- The answer is simply the difference in heights. Examiners award marks for the diagram/setup (1), correct use of tan for each plane (1), and the final subtraction (1).