Calculus Ladder problem, changing radians?

Question

A 30-foot ladder is leaning against a building. The base of the ladder is being pulled away from the building at a rate of 4 ft per second. How fast is the angle between the base of the ladder and the ground changing when the base is 10 ft away from the building?

I know that I have to use the derivative of a trig function, I believe cos because that lets you use the length of the ladder and the ground as it is changing.

cos = x over 30
-sin = 1 over 30

this is where I get stuck. Any advise as how to solve this problem would be greatly appreciated. thank you

Answer 1

If the base is x ft away from the building, the angle θ is given by cos θ = x/30.
Differentiating both with respect to time t, we get
-sin θ dθ/dt = 1/30 dx/dt.
Now when x = 10, cos θ = 1/3 and so sin θ = √(1-(1/3)^2) = √(8/9) = 2√2 / 3.
So we have
-2√2 / 3 (dθ/dt) = 1/30 (4)
=> dθ/dt = -4/30 . 3/(2√2)
= -1/(5√2)
≈ -0.141 radians/sec.