This problem can be solved using related rates in calculus. Here are the steps:

1. First, we need to understand the problem and the relationship between the variables. We have a right triangle formed by the kite, the point on the ground directly below the kite, and the person flying the kite. The hypotenuse of this triangle is the length of the string (200 ft), the height is the distance of the kite above the ground (100 ft), and the base is the horizontal distance the kite moves.

2. The angle θ we're interested in is between the string (hypotenuse) and the horizontal line (base). We can find this angle using the tangent function: tan(θ) = opposite/adjacent = height/base.

3. We're given that the height is constant (100 ft) and the base is increasing at a rate of 8 ft/s. We're asked to find dθ/dt, the rate at which the angle is decreasing, when the string is 200 ft long.

4. Using the Pythagorean theorem, we can find the base when the string is 200 ft long. We have 200² = 100² + base², which gives us base = sqrt(200² - 100²) = 100√3 ft.

5. Now we can find tan(θ) = 100/(100√3) = 1/√3. So, θ = arctan(1/√3).

6. We can differentiate both sides of the equation tan(θ) = height/base with respect to time t. This gives us sec²(θ) * dθ/dt = -height/(base²) * db/dt.

7. We can substitute the given values into this equation: 1/(cos²(θ)) * dθ/dt = -100/(100√3)² * 8.

8. We know that cos(θ) = adjacent/hypotenuse = 100√3/200 = √3/2. So, sec²(θ) = 1/(cos²(θ)) = 4/3.

9. Substituting these values into the equation gives us (4/3) * dθ/dt = -100/(30000) * 8.

10. Solving for dθ/dt gives us dθ/dt = -3/100 rad/s.

So, the angle between the string and the horizontal is decreasing at a rate of -3/100 rad/s when 200 ft of string have been let out.

Question

This problem can be solved using related rates in calculus. Here are the steps:

1. First, we need to understand the problem and the relationship between the variables. We have a right triangle formed by the kite, the point on the ground directly below the kite, and the person flying the kite. The hypotenuse of this triangle is the length of the string (200 ft), the height is the distance of the kite above the ground (100 ft), and the base is the horizontal distance the kite moves.

2. The angle θ we're interested in is between the string (hypotenuse) and the horizontal line (base). We can find this angle using the tangent function: tan(θ) = opposite/adjacent = height/base.

3. We're given that the height is constant (100 ft) and the base is increasing at a rate of 8 ft/s. We're asked to find dθ/dt, the rate at which the angle is decreasing, when the string is 200 ft long.

4. Using the Pythagorean theorem, we can find the base when the string is 200 ft long. We have 200² = 100² + base², which gives us base = sqrt(200² - 100²) = 100√3 ft.

5. Now we can find tan(θ) = 100/(100√3) = 1/√3. So, θ = arctan(1/√3).

6. We can differentiate both sides of the equation tan(θ) = height/base with respect to time t. This gives us sec²(θ) * dθ/dt = -height/(base²) * db/dt.

7. We can substitute the given values into this equation: 1/(cos²(θ)) * dθ/dt = -100/(100√3)² * 8.

8. We know that cos(θ) = adjacent/hypotenuse = 100√3/200 = √3/2. So, sec²(θ) = 1/(cos²(θ)) = 4/3.

9. Substituting these values into the equation gives us (4/3) * dθ/dt = -100/(30000) * 8.

10. Solving for dθ/dt gives us dθ/dt = -3/100 rad/s.

So, the angle between the string and the horizontal is decreasing at a rate of -3/100 rad/s when 200 ft of string have been let out.

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