Boat recedes from a tower — angle of depression changes from 45° to 30° in 5 s: A man at the top of a tower observes a boat. When the boat is 60 m from the base, the angle of depression is 45°. After 5 seconds, the angle of depression becomes 30°. Assuming straight-line motion in still water, what is the approximate speed of the boat (km/h)?

Introduction / Context:Changing angles of depression from the same height translate into changing horizontal distances. With height known from one sighting, the second sighting gives a farther horizontal distance. The difference over time gives horizontal speed.

Given Data / Assumptions:

• At first: horizontal distance x₁ = 60 m; angle = 45° ⇒ tower height h = x₁ × tan45° = 60 m.
• After 5 s: angle = 30° ⇒ horizontal distance x₂ = h / tan30° = 60√3 m.
• Straight-line motion away from the tower, still water.

Concept / Approach:Speed v = (x₂ − x₁)/time. Convert m/s to km/h by multiplying by 3.6.

Step-by-Step Solution:x₁ = 60 m; x₂ = 60√3 ≈ 103.92 mΔx ≈ 103.92 − 60 = 43.92 m in 5 s ⇒ v ≈ 43.92/5 = 8.784 m/sIn km/h: 8.784 × 3.6 ≈ 31.62 km/h ≈ 32 km/h (rounded)

Verification / Alternative check:Reversing: 32 km/h ≈ 8.888… m/s; over 5 s ≈ 44.44 m; close to Δx ≈ 43.92 m — consistent with rounding.

Why Other Options Are Wrong:36, 38, 42 km/h overshoot the computed horizontal speed; 30 km/h undershoots; 32 km/h is the nearest correct approximation.

Common Pitfalls:Using heights instead of horizontal distances; mixing degrees with radians; forgetting unit conversion to km/h.

Final Answer:32 Km/h