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).

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:

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