The base of a right triangle is fixed at 4 meters. The height y is increasing at a rate of 2m/s when y = 3 meters. Determine the rate at which theta is increasing, in radians per second, at this time.

To find the rate at which the angle \( \theta \) is increasing in a right triangle with a fixed base and a variable height, we can use trigonometry and related rates.

### Step-by-Step Solution:

1. **Identify Variables and Known Rates:**
   - Base of the triangle (\( x \)) is fixed at \( 4 \) meters.
   - Height (\( y \)) is variable and is increasing at \( \frac{dy}{dt} = 2 \, \text{m/s} \).
   - At the instant we are considering, \( y = 3 \) meters.

2. **Set Up the Trigonometric Relation:**
   - In a right triangle, we can express \( \tan(\theta) \) in terms of \( y \) and \( x \):
     \[
     \tan(\theta) = \frac{y}{x}
     \]
   - Since \( x \) is fixed, this relationship simplifies to \( \theta \) being a function of \( y \) alone.

3. **Differentiate with Respect to Time:**
   - We differentiate \( \tan(\theta) = \frac{y}{x} \) with respect to time \( t \):
     \[
     \sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{x} \cdot \frac{dy}{dt}
     \]
   - Solving for \( \frac{d\theta}{dt} \):
     \[
     \frac{d\theta}{dt} = \frac{1}{x \cdot \sec^2(\theta)} \cdot \frac{dy}{dt}
     \]

4. **Calculate \( \sec^2(\theta) \):**
   - Since \( \tan(\theta) = \frac{y}{x} = \frac{3}{4} \) at the given time, we can find \( \theta \) using the tangent inverse:
     \[
     \theta = \tan^{-1}\left(\frac{3}{4}\right)
     \]
   - Now, \( \sec^2(\theta) = 1 + \tan^2(\theta) \):
     \[
     \sec^2(\theta) = 1 + \left(\frac{3}{4}\right)^2 = 1 + \frac{9}{16} = \frac{25}{16}
     \]

5. **Substitute Values:**
   - Substitute \( x = 4 \), \( \sec^2(\theta) = \frac{25}{16} \), and \( \frac{dy}{dt} = 2 \) into the differentiated equation:
     \[
     \frac{d\theta}{dt} = \frac{1}{4 \cdot \frac{25}{16}} \cdot 2
     \]
     Simplifying:
     \[
     \frac{d\theta}{dt} = \frac{1}{\frac{100}{16}} \cdot 2 = \frac{16}{100} \cdot 2 = \frac{32}{100} = 0.32 \, \text{radians per second}
     \]

### Answer:
The rate at which \( \theta \) is increasing at this time is \( \frac{d\theta}{dt} = 0.32 \, \text{radians per second} \).

Would you like a breakdown of any specific step, or do you have additional questions?

Here are some related questions for further exploration:
1. How would the answer change if the base length was variable instead of fixed?
2. What if the rate of height increase was a function of time?
3. Can this problem be solved using a different trigonometric function?
4. How does the result change if the triangle’s base is extended to a larger fixed length?
5. What implications would this rate of angle change have in real-world applications like tilting structures?

**Tip:** When dealing with related rates, always differentiate with respect to time, even if the initial relationship doesn’t directly involve time.

This problem demonstrates how to calculate the rate at which the angle θ is increasing in a right triangle with a fixed base of 4 meters and a variable height. Using related rates and trigonometric identities, the solution finds that θ increases at 0.32 radians per second when the height is 3 meters and increasing at 2 m/s. Suitable for students studying trigonometry and calculus in Grades 11-12.

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