Answer
The angle between the string and the horizontal is decreasing at a rate of $~~\frac{1}{50}~rad/s$
Work Step by Step
Let $x$ be the horizontal distance between the string on the ground and the point directly below the kite. Let $\theta$ be the angle the string makes above the horizontal.
We can relate $x$ and $\theta$ in an equation. Then we can differentiate both sides with respect to $t$:
$\frac{x}{100} = cot~\theta$
$x = 100~cot~\theta$
$\frac{dx}{dt} = -100~csc^2~\theta~\frac{d\theta}{dt}$
$\frac{d\theta}{dt} = -\frac{sin^2~\theta}{100}\frac{dx}{dt}$
$\frac{d\theta}{dt} = -\frac{(\frac{100~ft}{200~ft})^2~}{100}(8~ft/s)$
$\frac{d\theta}{dt} = -\frac{8}{400}~rad/s$
$\frac{d\theta}{dt} = -\frac{1}{50}~rad/s$
The angle between the string and the horizontal is decreasing at a rate of $~~\frac{1}{50}~rad/s$
