A lighthouse stands tall on a cliff by the sea, watching over ships that pass by. One day a ship is seen approaching the shore and from the top of the lighthouse, the angles of depression of the ship are observed to be $30^\circ$ and $45^\circ$ as it moves from point P to point Q. The height of the lighthouse is 50 metres.
Generated by claude-sonnet-4-6 · 2026-06-15 10:34 · grounding stimulus
Model Answer
Let A be the top of the lighthouse, B be the base, height AB = 50 m.
(i) Distance of ship at Q (angle of depression = 45°):
In △ABQ: tan 45° = AB/BQ
$$1 = \frac{50}{BQ} \Rightarrow BQ = 50 \text{ m}$$
(ii) Measures of ∠PBA and ∠QBA:
Since AB ⊥ BP (base is horizontal), both ∠PBA and ∠QBA = 90°.
(iii) Distance travelled by the ship (PQ):
In △ABP: tan 30° = AB/BP
$$\frac{1}{\sqrt{3}} = \frac{50}{BP} \Rightarrow BP = 50\sqrt{3} \text{ m}$$
Distance travelled = PQ = BP − BQ
$$= 50\sqrt{3} - 50 = 50(\sqrt{3}-1) \text{ m} \approx 50(1.732-1) = 36.6 \text{ m}$$
Source: Chapter 9 – Some Applications of Trigonometry
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Explanation
- Part (i): Straightforward tan ratio with 45°; tan 45° = 1 so BQ = height = 50 m.
- Part (ii): ∠PBA and ∠QBA are the angles at base B in the respective right triangles, both equal 90° since the lighthouse is vertical and the ground is horizontal.
- Part (iii): Find BP using tan 30°, then subtract BQ from BP. Remember to simplify and rationalize. Examiners expect the expression $50(\sqrt{3}-1)$ m clearly shown.