A man atop a tower observes a boat moving away. Initially, the boat is 60 meters from the tower with a 45° angle of depression. After 5 seconds, the angle reduces to 30°. Calculate the boat's approximate speed in still water.

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Accepted Answer

**Step 1: Determine the tower height**  
When the boat is 60 meters away (point C), the angle of depression is 45°. Using the tangent function:  
$$
	an 45° = \frac{	ext{Height (AB)}}{	ext{Horizontal distance (AC)}} = \frac{h}{60}
$$  
Since $$	an 45° = 1$$:  
$$
h = 60 	ext{ meters}
$$

**Step 2: Calculate the new distance after 5 seconds**  
When the angle becomes 30° (point D):  
$$
	an 30° = \frac{60}{	ext{AD}} \implies \frac{1}{\sqrt{3}} = \frac{60}{	ext{AD}} \implies 	ext{AD} = 60\sqrt{3} 	ext{ meters}
$$

**Step 3: Compute the distance traveled**  
The boat moves from C to D:  
$$
	ext{Distance} = 60\sqrt{3} - 60 = 60(\sqrt{3} - 1) 	ext{ meters}
$$

**Step 4: Convert distance to speed**  
Speed in m/s:  
$$
	ext{Speed} = \frac{60(\sqrt{3} - 1)}{5} = 12(\sqrt{3} - 1) 	ext{ m/s}
$$  
Convert to km/h:  
$$
12(\sqrt{3} - 1) 	imes 3.6 \approx 12 	imes 0.732 	imes 3.6 \approx 31.5 	ext{ km/h} \approx \mathbf{32 	ext{ km/h}}
$$

**Final Answer:** The approximate speed of the boat is **32 km/h**

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