Related rates calculus problem involving shadow lengths A light on the ground is 30 feet away from a building. A 4 foot tall man is walking from the light to the building at a rate of 3 feet per second. He is casting a shadow on the side of the building. At what rate is his shadow shrinking when he is 5 feet from the building?
This is my figure: 
Is this correct?
How do I solve this problem? I tried using similar triangles but couldn't succeed. Could someone explain this with a figure?
Modified figure: light is on the ground.

 A: Perhaps you can do the drawing. Let $L$ be the location of the light. (It is on the ground!) Let $F$ be the location of the man's feet, and let $H$ be the top of his head. Let $B$ be the bottom of the building.  Draw the horizontal line $LB$. Draw the vertical line $FH$ to represent the short thin man. 
Draw the line through $L$ and $H$. Let it meet the wall of the building at $S$. Then $BS$ is the length of the shadow. 
It is convenient to call the distance $LF$ by the name $x$. We are told the man is walking at $3$ feet per second, so $\frac{dx}{dt}=3$.
Let $y=BS$. Note that triangles $LBS$ and $LFH$ are similar, so 
$$\frac{y}{30}=\frac{4}{x}.$$
We may want to rewrite this as
$$y=\frac{120}{x}.$$
Differentiate. We get
$$\frac{dy}{dt}=-\frac{120}{x^2}\frac{dx}{dt}.$$
Now it is almost over. 
A: My approach would be to define a function which gives us the shadow height (S) in dependence of his walked distance (x):
x/4 = 30/S -> S(x) = 120/x
Now we know that x(t) = 3*t -> S(t)= 40/t.
All you have to do now is to find S'(25)!
but don't take my word for it and ask if something is unclear.
