The picture below corresponds to the problem described.

∣CD∣ is the height of the man. ∣CD∣=6ft. ∣EB∣ is the height of the lamp post. ∣EB∣=18ft. Denote ∣DE∣=x. In order to find ∣AD∣, we write the definition of tan(α), where α=<BAE. We get: tan(α)=∣AD∣∣CD∣=∣AE∣∣BE∣. ∣AE∣=∣AD∣+∣DE∣=∣AD∣+x. We receive: 6(∣AD∣+x)=18∣AD∣. We get: ∣AD∣=2x. Suppose that a man walked a distance ΔS=vΔt, where Δt is a small period of time and v=4ft/sec. We receive new segment C′D′ of the length 6ft. and a new point A′, which is on the line BC′. ∣D′E∣=∣DE∣+vΔt=∣DE∣+4Δt=x+4Δt. Denote: α′=<BA′E. We get: tan(α′)=∣A′D′∣∣C′D′∣=∣A′D′∣+∣D′E∣∣BE∣. From the latter we find: 6(∣A′D′∣+∣D′E∣)=18∣A′D′∣. Thus, ∣A′D′∣=21∣D′E∣=21(x+4Δt). ∣AD∣−∣A′D′∣=2Δt. From the latter we get that the shadow lengthen with a speed 2ft/sec.
Answer: the shadow lengthen with the speed 2ft/sec.
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