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To solve this problem, let’s define the key variables:

1. Let \( h = 100 \) feet (the height of the kite above the ground).
2. The kite is moving horizontally at a speed of \( \frac{dx}{dt} = 4 \) ft/s.
3. Let \( s \) be the length of the string let out, which is given as 150 ft.
4. Let \( \theta \) be the angle between the string and the horizontal line.

We are asked to find the rate at which this angle \( \theta \) is decreasing. This corresponds to finding \( \frac{d\theta}{dt} \) when \( s = 150 \) ft.

### Step 1: Set Up the Relationship
Using trigonometry, we know:
\[
\cos \theta = \frac{x}{s}
\]
where \( x \) is the horizontal distance from the person holding the kite to the point directly under the kite. Also, since \( x \) and \( h \) form a right triangle with hypotenuse \( s \), we have:
\[
s^2 = x^2 + h^2
\]

### Step 2: Differentiate with Respect to Time
Differentiate both sides of \( s^2 = x^2 + h^2 \) with respect to \( t \):
\[
2s \frac{ds}{dt} = 2x \frac{dx}{dt}
\]
Since the length of the string is constant (\( s = 150 \) ft), \( \frac{ds}{dt} = 0 \). Simplifying, we get:
\[
0 = 2x \frac{dx}{dt}
\]
So

A kite 100 ft above the ground moves horizontally at a speed of 4 ft/s. At what rate is the angle between the string and the horizontal decreasing when 150 ft of string has been let out?

Find the rate at which the angle between the string of a kite 100 ft above the ground and the horizontal decreases as it moves at 4 ft/s. This problem uses trigonometry, related rates, and the Pythagorean theorem to determine the answer.

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