We can relate the horizontal extension x of the string to the angle of elevation θ: tanθ=100/x. Differentiate with respect to time t: sec²θdθ/dt=-(100/x²)dx/dt.
dθ/dt is the rate of change of the angle of elevation, and dx/dt is the horizontal rate of change of the length of the string, which is 10 ft/s. Therefore sec²θdθ/dt=-(100/x²)(10)=-1000/x².
When θ=30°, x=100cotθ=100√3 feet (this is the horizontal ground distance to a point directly beneath the kite, not the length of the kite string). We plug in (x,θ)=(100√3,30):
sec²θdθ/dt=1000/(100√3)²=-1/30, sec²30=4/3, and dθ/dt=-1/30×¾=-1/40=-0.025°/s approx. This means that the rate of change is a decrease in the angle of elevation which is 0.025 degrees per second at the instantaneous point where the angle of elevation was 30°. At this point, the length of the string was 200 feet (at an angle of 30°), but the string is extended horizontally, not at an angle of 30°.
Alternative solution using x=100cotθ, dx/dt=-100cosec²θdθ/dt.
dx/dt=10, θ=30°, 10=-100×4dθ/dt, so dθ/dt=-1/40 degrees per second.