From the top of a 45 m high light house, the angles of depression of two ships, on the opposite side of it, are observed to be 30° and 60°. If the line joining the ships passes through the foot of the light house, find the distance between the ships. (Use $\sqrt{3} = 1.73$)
Generated by claude-sonnet-4-6 · 2026-06-15 10:34 · grounding rag
Model Answer
Diagram: Let AB = 45 m be the lighthouse. Ships are at C and D on opposite sides, with foot of lighthouse at B.
Angle of depression to C = 60°, angle of depression to D = 30°.
For Ship C (angle of depression = 60°):
In △ABC,
$$\tan 60° = \frac{AB}{BC}$$
$$\sqrt{3} = \frac{45}{BC}$$
$$BC = \frac{45}{\sqrt{3}} = \frac{45\sqrt{3}}{3} = 15\sqrt{3} \text{ m}$$
For Ship D (angle of depression = 30°):
In △ABD,
$$\tan 30° = \frac{AB}{BD}$$
$$\frac{1}{\sqrt{3}} = \frac{45}{BD}$$
$$BD = 45\sqrt{3} \text{ m}$$
Distance between the two ships:
$$CD = BC + BD = 15\sqrt{3} + 45\sqrt{3} = 60\sqrt{3} \text{ m}$$
$$= 60 \times 1.73 = \boxed{103.8 \text{ m}}$$
Source: Some Applications of Trigonometry, Chapter 9
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Explanation
- Since the ships are on opposite sides, distances BC and BD are added (not subtracted).
- The angle of depression from the top equals the angle of elevation from the ship's position (alternate angles with horizontal), so directly use tan in the right triangle formed.
- Always rationalize $\frac{45}{\sqrt{3}}$ to get $15\sqrt{3}$ — examiners check this step.
- Substitute $\sqrt{3} = 1.73$ only in the final step to avoid rounding errors midway.
- Drawing a neat labelled diagram typically earns 1 mark on its own.