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The image shows a problem involving a kite gliding horizontally while its string is unspooled. The key elements of the problem are:

- The kite is at an altitude of 30 meters.
- The kite's horizontal velocity is 3 m/s.
- We are tasked with determining how fast the angle \(\theta\) between the kite string and the horizontal decreases when 60 meters of string have been unspooled.

The image also contains parts of the problem that involve:
- (a) Expressing \(\theta\) as a function of the kite's horizontal position \(x\).
- (b) Calculating the value of \(x\) when 60 meters of string is unspooled.
- (c) Finding the rate at which the angle \(\theta\) is decreasing, \(\frac{d\theta}{dt}\), given that the horizontal velocity is 3 m/s.

Would you like detailed steps and explanations for each part of this problem? 

Here are five related questions to extend this:

1. How is the horizontal position \(x\) related to the total length of the string?
2. Why is \(\theta = \arccos\left(\frac{x}{60}\right)\) the appropriate expression for the angle?
3. What is the significance of using the derivative \(\frac{d\theta}{dt}\) in this context?
4. How do changes in the horizontal velocity of the kite affect \(\frac{d\theta}{dt}\)?
5. Can you apply this concept to determine the rate of change of the string length as the kite moves?

**Tip**: Always ensure your units are consistent (e.g., seconds, meters) when dealing with rates of change like velocities and angular velocities.

A kite glides horizontally at an altitude of 30 meters while we unspool the string. We want to determine the rate at which the angle between the string and the horizon decreases after 60 meters of string has been unspooled. Given: the kite's horizontal velocity is 3 m/s.

Learn how to calculate the rate of change of the angle between a kite's string and the horizontal after 60 meters of string have been unspooled. The problem involves trigonometric functions, Pythagorean Theorem, and implicit differentiation, suitable for students in Grades 10-12.

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