From the top of a tower, a boat is observed moving directly away. When the horizontal distance is 60 m, the angle of depression is 45°. After 5 seconds, the angle of depression becomes 30°. Assuming straight-line motion on still water, find the.

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Introduction / Context:Angles of depression equal angles of elevation from the boat. With two angles at two horizontal distances separated by a known time, we can deduce tower height and the boat’s horizontal speed. Given Data / Assumptions: First horizontal distance d1 = 60 m with angle = 45°. After 5 s, angle = 30° ⇒ new horizontal distance d2 unknown. Right-triangle geometry, flat ground. Concept / Approach:tan θ = height / horizontal_distance. At 45°, height equals horizontal distance. Use that to get tower height, then find d2 from 30°, then speed = (d2 − d1)/time. Step-by-Step Solution: h = d1 * tan 45° = 60 m.At 30°: tan 30° = h/d2 ⇒ 1/√3 = 60/d2 ⇒ d2 = 60√3 ≈ 103.92 m.Distance gained in 5 s = 103.92 − 60 = 43.92 m.Speed = 43.92/5 ≈ 8.784 m/s ⇒ 8.784 * 3.6 ≈ 31.5 km/h. Verification / Alternative check:Recomputing back to angles with this speed gives the same distances and angles in 5 s; numerics align. Why Other Options Are Wrong:30, 33, 34, 28.8 km/h correspond to rounding extremes or arithmetic slips when converting to km/h. Common Pitfalls:Using d2 − d1 with wrong sign; mixing degrees and radians; skipping the conversion from m/s to km/h (multiply by 3.6). Final Answer:31.5 km/h

From the top of a tower, a boat is observed moving directly away. When the horizontal distance is 60 m, the angle of depression is 45°. After 5 seconds, the angle of depression becomes 30°. Assuming straight-line motion on still water, find the boat’s speed (km/h).

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