Two Boats, One Lighthouse: Angles of Depression on the Same Side
Two fishing boats are anchored in line with a lighthouse on the Odisha coast, both on the same side of it, and the lamp is 20 m above sea level. From the lamp, an observer measures the angles of depression of the nearer and the farther boat as 45° and 30° respectively. A trainee at the harbour office works out the gap between the boats by adding the two horizontal distances from the lighthouse and reports it as 20(√3 + 1) m; the harbour master wants this checked before a dredging permit is issued for the channel between the boats. Separately, the coast guard asks whether a 30 m rescue rope lowered straight from the lamp would reach the nearer boat, allowing for the rope being pulled taut. All observations lie in the same vertical plane, the sea is calm, and the angles may be treated as exact.
1. Find the horizontal distance of each boat from the foot of the lighthouse and hence the actual gap between the boats.
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The angle of depression from the lamp to a boat equals the angle of elevation from the boat to the lamp (alternate angles between the horizontal lines). So use right triangles with height 20 m.
Nearer boat: tan 45° = 20/d₁ → 1 = 20/d₁ → d₁ = 20 m.
Farther boat: tan 30° = 20/d₂ → 1/√3 = 20/d₂ → d₂ = 20√3 ≈ 34.64 m.
Because both boats are on the same side, the gap is the difference:
20√3 − 20 = 20(√3 − 1) ≈ 20 × 0.732 = 14.64 m.
So the channel to be dredged is about 14.6 m wide, not the larger figure the trainee reported.
2. Explain why the trainee's answer of 20(√3 + 1) m is wrong, and say when that formula would be correct.
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The trainee has added the distances instead of subtracting them: 20 + 20√3 = 20(1 + √3) ≈ 54.64 m.
Addition is correct only when the two objects lie on opposite sides of the foot of the tower, because then the gap equals d₁ + d₂. Here the problem states both boats are on the same side of the lighthouse, so the farther boat's distance already includes the nearer boat's distance, and the gap is d₂ − d₁.
So the correct value is 20(√3 − 1) ≈ 14.64 m, and the trainee's figure overstates the channel width by about 40 m.
Practical consequence: a permit based on 54.64 m would authorise dredging far beyond the actual channel, wasting money and possibly disturbing the berth of the nearer boat.
3. Would a 30 m rope lowered straight from the lamp reach the nearer boat? Justify with a calculation.
Read the worked answer
The rope runs along the line from the lamp to the boat, which is the hypotenuse of the right triangle with height 20 m and horizontal base 20 m.
Length = √(20² + 20²) = √800 = 20√2 ≈ 20 × 1.414 = 28.28 m.
Since 28.28 m < 30 m, the rope does reach the nearer boat, with about 1.7 m to spare.
Check that this is not confused with the horizontal distance (20 m) or the height (20 m): the straight line from the lamp to the boat is longer than either. For the farther boat the same calculation would give √(20² + (20√3)²) = √(400 + 1200) = 40 m, so the 30 m rope would not reach that boat.