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This problem involves a kite gliding horizontally while unspooling a string, and we need to determine how the angle between the string and the horizontal decreases over time.

### Let's break down each part of the problem:

#### Given information:
- Altitude \( h = 30 \) m (the height of the kite).
- The string is being unspooled such that 60 meters of string have been unspooled at a specific moment.
- The kite's horizontal velocity is \( \frac{dx}{dt} = 3 \text{ m/s} \), where \( x \) is the horizontal distance of the kite from the point where the string is attached to the ground.

We need to find the angle \( \theta \) between the string and the horizontal line at that moment, as well as the rate of change of this angle, \( \frac{d\theta}{dt} \).

---

### (a) Expressing \( \theta \) as a function of \( x \):

From the problem, we have a right triangle formed by:
- The height \( h = 30 \) meters (vertical side).
- The horizontal distance \( x \) meters (horizontal side).
- The length of the string \( L = 60 \) meters (the hypotenuse of the triangle).

From basic trigonometry:
\[
\cos(\theta) = \frac{x}{L}
\]
Thus, the angle \( \theta \) can be written as:
\[
\theta = \cos^{-1} \left( \frac{x}{L} \right)
\]
Substituting \( L = 60 \) m:
\[
\theta = \cos^{-1} \left( \frac{x}{60} \right)
\]

---

### (b) Finding the value of \( x \):

At the moment in question, we can apply the Pythagorean theorem to solve for \( x \). The relationship between the height, the horizontal distance, and the length of the string is:
\[
x^2 + h^2 = L^2
\]
Substituting \( h = 30 \) m and \( L = 60 \) m:
\[
x^2 + 30^2 = 60^2
\]
\[
x^2 + 900 = 3600
\]
\[
x^2 = 2700
\]
\[
x = \sqrt{2700} = 30\sqrt{3} \, \text{m}
\]

---

### (c) Finding \( \frac{d\theta}{dt} \):

To find the rate of change of \( \theta \), we implicitly differentiate the trigonometric relationship \( \cos(\theta) = \frac{x}{L} \) with respect to time \( t \). We get:
\[
-\sin(\theta) \frac{d\theta}{dt} = \frac{1}{L} \frac{dx}{dt}
\]
Solving for \( \frac{d\theta}{dt} \):
\[
\frac{d\theta}{dt} = -\frac{1}{L \sin(\theta)} \frac{dx}{dt}
\]
Now, we substitute known values:
- \( L = 60 \) m,
- \( \frac{dx}{dt} = 3 \) m/s,
- \( \sin(\theta) = \frac{h}{L} = \frac{30}{60} = \frac{1}{2} \).

Thus:
\[
\frac{d\theta}{dt} = -\frac{1}{60 \times \frac{1}{2}} \times 3 = -\frac{1}{30} \times 3 = -\frac{1}{10} \, \text{rad/s}
\]

---

### Summary of results:
(a) \( \theta = \cos^{-1} \left( \frac{x}{60} \right) \).

(b) \( x = 30\sqrt{3} \, \text{m} \).

(c) \( \frac{d\theta}{dt} = -\frac{1}{10} \, \text{rad/s} \).

Would you like further clarification on any part of this solution? 

Here are 5 related questions you could explore next:
1. How would the solution change if the altitude of the kite were different?
2. How do different rates of horizontal velocity affect \( \frac{d\theta}{dt} \)?
3. What would happen if the string length increased faster than it currently does?
4. How would the trigonometric relationships change if the kite's altitude varied over time?
5. How could you solve this problem if the kite’s trajectory were not horizontal?

### Tip:
Whenever you're working with rates of change (like \( \frac{d\theta}{dt} \)), implicit differentiation is a powerful tool for connecting related variables!

A kite glides horizontally at an altitude of 30 m while we unspool the string. We need to determine the rate at which the angle between the string and the horizontal decreases over time once 60 m of string has been unspooled. The kite's horizontal velocity at that instant is 3 m/s.

Learn how to solve a trigonometry-based problem involving a kite gliding horizontally at a fixed altitude of 30 meters. Explore key mathematical concepts like implicit differentiation and the Pythagorean theorem to determine the rate of change of the angle between the string and the horizontal line when 60 meters of string is unspooled. Suitable for high school students in grades 10-12.

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