so hmm notice the picture
at the moment r = 200, or the string is 200ft to the kite up above, at that moment, those are the values for "x" and θ
however, "r" has a rate, so.. we wont' be using that one to get the angle, we'll use the "constant y" or 100
thus [tex]\bf tan(\theta)=\cfrac{100}{x}\implies tan(\theta)=100x^{-1}\\\\
-----------------------------\\\\
sec^2(\theta)\cfrac{d\theta}{dt}=-\cfrac{100}{x^2}\cdot \cfrac{dx}{dt}\implies \cfrac{1}{cos^2(\theta)}\cdot\cfrac{d\theta}{dt}=-\cfrac{100}{x^2}\cdot \cfrac{dx}{dt}
\\\\\\
\cfrac{d\theta}{dt}=\cfrac{-cos^2(\theta)\cdot 100\cdot \frac{dx}{dt}}{x^2}[/tex]
so hmm, pretty sure you can take from there, keeping in mind that dx/dt is 8
