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This is a related rates problem involving trigonometry.

### Given:
- The observer is 5 miles from the lift-off point.
- The angle \(\theta = \frac{\pi}{5}\) radians at a particular moment.
- The rate of change of the angle is \(\frac{d\theta}{dt} = 0.1\) radians/minute.

We need to find how fast the balloon is rising, which corresponds to \(\frac{dy}{dt}\), where \(y\) is the height of the balloon.

### Step-by-step solution:

1. **Relationship between the variables:**
   Since the balloon is rising vertically, we can use the tangent function:
   \[
   \tan(\theta) = \frac{y}{5}
   \]
   where \(y\) is the height of the balloon, and 5 is the horizontal distance from the observer to the lift-off point.

2. **Differentiate with respect to time \(t\):**
   Using implicit differentiation:
   \[
   \frac{d}{dt} \left( \tan(\theta) \right) = \frac{d}{dt} \left( \frac{y}{5} \right)
   \]
   The derivative of \(\tan(\theta)\) is \(\sec^2(\theta) \cdot \frac{d\theta}{dt}\), and the derivative of \(\frac{y}{5}\) is \(\frac{1}{5} \cdot \frac{dy}{dt}\):
   \[
   \sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{5} \cdot \frac{dy}{dt}
   \]

3. **Solve for \(\frac{dy}{dt}\):**
   \[
   \frac{dy}{dt} = 5 \cdot \sec^2(\theta) \cdot \frac{d\theta}{dt}
   \]

4. **Find \(\sec^2(\theta)\):**
   We know \(\theta = \frac{\pi}{5}\), so:
   \[
   \sec(\theta) = \frac{1}{\cos(\frac{\pi}{5})}
   \]
   From trigonometric tables or a calculator, \(\cos\left(\frac{\pi}{5}\right) \approx 0.809\), so:
   \[
   \sec(\theta) \approx \frac{1}{0.809} \approx 1.236
   \]
   Thus:
   \[
   \sec^2(\theta) \approx 1.236^2 \approx 1.528
   \]

5. **Substitute values:**
   Now, substitute \(\sec^2(\theta) \approx 1.528\) and \(\frac{d\theta}{dt} = 0.1\) into the equation:
   \[
   \frac{dy}{dt} = 5 \cdot 1.528 \cdot 0.1 = 0.764 \, \text{miles per minute}
   \]

### Final Answer:
The balloon is rising at approximately **0.764 miles per minute**.

Would you like a deeper explanation on any part of the solution?

Here are 5 related questions you might find interesting:
1. How would the result change if the observer were located 10 miles from the lift-off point?
2. How does the rate of change of the balloon's height depend on the angle \(\theta\)?
3. Can you derive the formula for the rate of change of the balloon's height using a different trigonometric function?
4. What would happen if the rate of change of the angle, \(\frac{d\theta}{dt}\), increased to 0.2 rad/min?
5. How would this problem change if the balloon were moving along a diagonal path instead of vertically?

**Tip:** Always visualize related rates problems by sketching a diagram and labeling all the variables!

A hot air balloon rising vertically is tracked by an observer located 5 miles from the lift-off point. At a certain moment, the angle between the observer's line of sight and the horizontal is π/5, and it is changing at a rate of 0.1 rad/min. How fast is the balloon rising at this moment?

Learn how to solve a related rates problem involving a hot air balloon rising vertically, tracked by an observer 5 miles away. This problem utilizes trigonometric concepts and implicit differentiation to find the balloon's vertical speed. Suitable for Grades 11-12.

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