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An observer on top of a mountain 500 m high observes the angle of depression of two boats on the same side of the mountain to be 45° and 30°. The distance between the boats is:

  • A.\(500(\sqrt{3}-1)\) m
  • B.\(500\sqrt{3}\) m
  • C.\(500(\sqrt{3}+1)\) m
  • D.500 m

Answer: A

Let the mountain be at point P, with height PQ = 500m. Let the boats be A and B.

For the farther boat (B), angle of elevation is 30°. Let its distance be d₂. \(\tan 30° = \frac{500}{d_2} \Rightarrow d_2 = 500\sqrt{3}\).

For the nearer boat (A), angle of elevation is 45°. Let its distance be d₁. \(\tan 45° = \frac{500}{d_1} \Rightarrow d_1 = 500\).

The distance between the boats is \(d_2 - d_1 = 500\sqrt{3} - 500 = 500(\sqrt{3}-1)\) m.

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