As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 60°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships. (Use $\sqrt{3} = 1{\cdot}73$)
Generated by claude-sonnet-4-6 · 2026-06-15 10:35 · grounding rag
Model Answer
Diagram: Let AB = 75 m be the lighthouse. Let C and D be the two ships, with D farther from the lighthouse. Angles of depression to D and C are 30° and 60° respectively.
In right △ABC (nearer ship, angle of depression = 60°):
$$\tan 60° = \frac{AB}{BC}$$
$$\sqrt{3} = \frac{75}{BC} \implies BC = \frac{75}{\sqrt{3}} = 25\sqrt{3} \text{ m}$$
In right △ABD (farther ship, angle of depression = 30°):
$$\tan 30° = \frac{AB}{BD}$$
$$\frac{1}{\sqrt{3}} = \frac{75}{BD} \implies BD = 75\sqrt{3} \text{ m}$$
Distance between the two ships:
$$CD = BD - BC = 75\sqrt{3} - 25\sqrt{3} = 50\sqrt{3} \text{ m}$$
$$= 50 \times 1.73 = \boxed{86.5 \text{ m}}$$
Source: Chapter 9, Heights and Distances (Exercise 9.1, Q.13)
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Explanation
- The angle of depression from the top of the lighthouse equals the angle of elevation from the ship to the top (alternate angles), so tan is applied directly with height = 75 m.
- The ship closer to the lighthouse makes a larger angle of depression (60°), and the farther ship makes a smaller angle (30°) — a common point students get wrong.
- CD = BD − BC is the key final step.
- Always substitute √3 = 1.73 at the end and state the unit (metres).