a kite is 20 metres above the ground and is moving horizontally at a constant rate of 0.5 metres per second as the string is let out. at what rate is the angle of inclination of the string to the horizontal ground changing when 50 metres of string has been let out?

After t seconds, let θ radians be the angle the string makes with the horizontal ground, and let the projection of the string on the ground be x metres.
Given the horizontal speed of the kite is dx/dt=0.5m/s
To find dθ/dt when the length of string is 50 metres.
Since dθ/dt=(dx/dt)(dθ/dx)
tanθ=20/x--->x=20cotθ
Thus dx/dθ=-20/sin²θ--->dθ/dx=-sin²θ/20
Gives dθ/dt=-sin²θ/40
Need to find θ when the length of string let out is 50 metres.
sinθ=20/50
dθ/dt=-(20/50)²/40=-0.004radians/second.
I hope you understand.