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The problem in the image is a classic related rates problem, involving a right triangle formed by a ladder leaning against a wall. Here's a step-by-step breakdown:

### Given:
- The ladder's length \( L = 25 \) feet (constant).
- The base of the ladder is being pulled away from the wall at a rate of \( \frac{da}{dt} = 2 \) feet per second.
- When the base is \( a = 7 \) feet from the wall, we need to find the rate at which the angle \( \theta \) between the ladder and the wall is changing, i.e., \( \frac{d\theta}{dt} \).

We know the relationship between the ladder's position and the angle \( \theta \) using basic trigonometry:
\[
\cos(\theta) = \frac{b}{L}
\]
where \( b \) is the vertical distance from the ground to the top of the ladder, and \( L \) is the length of the ladder.

### Step 1: Use Pythagoras' theorem to relate \( a \), \( b \), and \( L \):
\[
a^2 + b^2 = L^2
\]
Since \( L = 25 \) feet:
\[
a^2 + b^2 = 25^2
\]
When \( a = 7 \):
\[
7^2 + b^2 = 625 \quad \Rightarrow \quad 49 + b^2 = 625 \quad \Rightarrow \quad b^2 = 576 \quad \Rightarrow \quad b = \sqrt{576} = 24 \text{ feet.}
\]

### Step 2: Differentiate the trigonometric equation with respect to time:
\[
\cos(\theta) = \frac{b}{L}
\]
Since \( L \) is constant, we have:
\[
\frac{d}{dt} [\cos(\theta)] = \frac{d}{dt} \left(\frac{b}{L}\right) = \frac{1}{L} \cdot \frac{db}{dt}
\]
Using the chain rule on the left-hand side:
\[
-\sin(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{L} \cdot \frac{db}{dt}
\]
Thus:
\[
\frac{d\theta}{dt} = -\frac{1}{L \sin(\theta)} \cdot \frac{db}{dt}
\]

### Step 3: Relate \( \sin(\theta) \) and use the related rates for \( \frac{db}{dt} \):
We know from Pythagoras' theorem that:
\[
\sin(\theta) = \frac{a}{L}
\]
So:
\[
\sin(\theta) = \frac{7}{25}
\]

Next, to find \( \frac{db}{dt} \), we differentiate the Pythagorean equation \( a^2 + b^2 = L^2 \) with respect to time \( t \):
\[
2a \cdot \frac{da}{dt} + 2b \cdot \frac{db}{dt} = 0
\]
Solving for \( \frac{db}{dt} \):
\[
\frac{db}{dt} = -\frac{a}{b} \cdot \frac{da}{dt} = -\frac{7}{24} \cdot 2 = -\frac{14}{24} = -\frac{7}{12} \text{ feet per second.}
\]

### Step 4: Substitute the values into the equation for \( \frac{d\theta}{dt} \):
\[
\frac{d\theta}{dt} = -\frac{1}{25 \cdot \frac{7}{25}} \cdot \left(-\frac{7}{12}\right) = \frac{1}{7} \cdot \frac{7}{12} = \frac{1}{12} \text{ radians per second.}
\]

Thus, the rate at which the angle \( \theta \) is changing when the base of the ladder is 7 feet from the wall is approximately **0.083 radians per second**.

Would you like a further breakdown or have any questions? 

Here are five related questions you can explore:
1. How does \( \frac{d\theta}{dt} \) change if the base is moving away faster?
2. How would the problem change if the ladder length were different?
3. What happens if the base is moving toward the wall instead of away?
4. Can you solve this using a different trigonometric function like \( \sin(\theta) \)?
5. How does the height \( b \) change over time as the base moves away?

**Tip:** Always differentiate implicitly when working with related rates, and remember to include the rates of change (derivatives) of all variables involved.

A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second. Find the rate at which the angle between the ladder and the wall of the house is changing when the base is 7 feet from the wall.

Solve a classic related rates problem: a 25-foot ladder leaning against a wall is pulled away at 2 feet per second. Learn how to calculate the rate of change of the angle when the base is 7 feet from the wall using trigonometric identities and implicit differentiation.

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