Question

A kite 100 ft above the ground moves horizontally at a speed of 3 ft/s. At what rate is the angle (in radians) between the string and the horizontal decreasing when 200 ft of string have been let out?
rad/s

Answer 1

Draw a diagram. It is clear that the distance z between the kite and the spot directly over its anchor point obeys
100/z = tanθ
when 200 ft of string have been let out,
sinθ = 100/200 so θ = π/6
and z = 100√3
Now, we have
-100/z^2 dz/dt = sec^2θ dθ/dt
-100/300 = 4 dθ/dt
dθ/dt = -1/12 rad/s

Answer 2

At a given time of t seconds, let the horizontal distance of the kite be x ft
and let the angle it makes with the ground be θ
let the length of string be d ft
given: dx/dt = 3 ft/s
find: dθ/dt when d = 200 ft

then
tanθ = 200/x
sec^ θ dθ/dt = -200/x^2 dx/dt

Your specific case:
when d = 200, sinθ = 100/200 = 1/2, and θ = 30°
tan30° = 200/x
x = 200/tan30 = 200/(√3/2) = 400/√3
also:
cosθ = √3/2
cos^2 θ = 3/4
sec^2 θ = 4/3 , so in

sec^ θ dθ/dt = -200/x^2 dx/dt
4/3 dθ/dt = -200/(400/√3)^2 * 3

your turn to do some arithmetic, solve for dθ/dt

(your answer will be in radians/sec, it did not matter that I used 30° or π/6 radians in the above calculations, but the end results is in radians because
the derivative rules for differentiating trig functions are valid only for radians)