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This problem involves related rates, specifically the rate at which the angle \(\theta\) between the fishing line and the water is changing as the fish is reeled in.

### Step-by-Step Solution:

1. **Identify the variables:**
   - Let \(x\) be the horizontal distance between the fisherman and the fish.
   - The height of the fishing rod above the water is constant at 10 feet.
   - The length of the fishing line is \(L = 25\) feet when the question asks for the rate of change of the angle.
   - The rate at which the fishing line is being reeled in is \(\frac{dL}{dt} = -1\) ft/sec (negative because the line length is decreasing).

2. **Establish a relationship:**
   - From the right triangle in the diagram, we can use the Pythagorean theorem:
     \[
     L^2 = x^2 + 10^2
     \]
     \[
     L^2 = x^2 + 100
     \]

3. **Differentiate with respect to time \(t\):**
   \[
   2L \frac{dL}{dt} = 2x \frac{dx}{dt}
   \]
   Simplifying:
   \[
   L \frac{dL}{dt} = x \frac{dx}{dt}
   \]

4. **Use trigonometry for the angle \(\theta\):**
   The tangent of the angle \(\theta\) is:
   \[
   \tan(\theta) = \frac{10}{x}
   \]
   Taking the derivative of both sides with respect to time \(t\):
   \[
   \sec^2(\theta) \frac{d\theta}{dt} = \frac{-10}{x^2} \frac{dx}{dt}
   \]
   Solving for \(\frac{d\theta}{dt}\):
   \[
   \frac{d\theta}{dt} = \frac{-10}{x^2} \cdot \frac{dx}{dt} \cdot \cos^2(\theta)
   \]

5. **Substitute known values:**
   - \(L = 25\) feet.
   - \(\frac{dL}{dt} = -1\) ft/sec.
   - Use \(L^2 = x^2 + 100\) to find \(x\) when \(L = 25\):
     \[
     25^2 = x^2 + 100
     \]
     \[
     625 = x^2 + 100 \implies x^2 = 525 \implies x = \sqrt{525} \approx 22.91 \, \text{ft}
     \]
   
   Now, substitute back into the equation to find \(\frac{dx}{dt}\):
   \[
   25(-1) = 22.91 \cdot \frac{dx}{dt}
   \]
   \[
   \frac{dx}{dt} = \frac{-25}{22.91} \approx -1.09 \, \text{ft/sec}
   \]

6. **Final step – Find \(\frac{d\theta}{dt}\):**
   Now use the tangent relation:
   \[
   \tan(\theta) = \frac{10}{22.91} \approx 0.436
   \]
   So, \(\theta \approx \tan^{-1}(0.436)\), which gives \(\theta \approx 23.58^\circ\).

   Finally, using the differentiated trigonometric relation:
   \[
   \frac{d\theta}{dt} \approx \frac{-10}{(22.91)^2} \cdot (-1.09) \cdot \sec^2(23.58^\circ)
   \]
   We know \(\sec^2(\theta) = 1 + \tan^2(\theta)\), so:
   \[
   \sec^2(23.58^\circ) \approx 1 + 0.436^2 = 1.19
   \]
   \[
   \frac{d\theta}{dt} \approx \frac{-10}{525} \cdot (-1.09) \cdot 1.19
   \]
   Simplifying:
   \[
   \frac{d\theta}{dt} \approx 0.00248 \, \text{rad/sec}
   \]

Thus, the angle \(\theta\) is changing at approximately **0.00248 radians per second**.

---

Would you like further clarification on any steps, or more details about related rates?

Here are 5 related questions:
1. How would the rate of change in the angle \(\theta\) differ if the line length changed at a different rate?
2. Can we solve a similar problem if the fisherman is on a boat at sea level?
3. What happens to \(\frac{d\theta}{dt}\) as the length of the line decreases?
4. How does the height of the fishing rod impact the rate of change of the angle?
5. Can we model a similar problem if the fish swims horizontally rather than being reeled in?

**Tip:** Always keep track of signs (positive or negative) in related rates problems, as they indicate whether quantities are increasing or decreasing!

A fish is reeled in at a rate of 1 foot per second from a point 10 feet above the water. At what rate is the angle θ between the line and the water changing when there is a total of 25 feet of line from the end of the rod to the water?

Learn how to find the rate of change of the angle θ as a fish is reeled in from a point 10 feet above the water. This related rates problem involves using the Pythagorean theorem, trigonometric relationships, and calculus. Detailed step-by-step solution provided, suitable for college-level students studying calculus.

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