Related rates differentiation word problem I am suppose to work this out somehow but I am not sure. I have tried many ways and I can't progress any further.
A streethg light his mounted at the top of a 15 ft tall pole. A man 6 ft tall walks away from the pole at a speed of 5ft/s along a straight path. How fast is the tip of the shadow moving when he is 40tf from the pole?
From this I know that I have a triangle with 2 known values on it, 15tf tall and then an opposite side of that is 6 feet tall. The rate of change of the bottom of the triangle is 5. I need to solve for c prime I am pretty sure. I am just not sure how to get that. I tried many different ways and none are correct.I tried to set up the pythagorean theorem but it didn't work and I am not sure why. I know that if I have 2 sides of a triangle I can figure out a third, so I tried to find the derivative of that but that didn't work.
 A: The $15$ foot tall lightpole and the $6$ foot tall man aren’t sides of the same triangle. You actually have two triangles. Let $A$ be the point at the base of the lightpole, $B$ the point at the top of the lightpole, $C$ the point where the man is standing, $D$ the top of his head, and $E$ the end of his shadow. Then you have a right triangle $ABE$ with a vertical side of $15$ feet and another right triangle $CDE$ with a vertical side of $6$ feet. These triangles are related in a rather special way: they’re _______?
Now let $x$ be the distance of the man from the lightpole, i.e., the length of $\overline{AC}$. You know what $dx/dt$ is. Let $y$ be the length of $\overline{AE}$, the distance from the lightpole to the tip of his shadow. You want to find $dy/dt$. If you can answer the question above, you should be able to find a relationship between $x$ and $y$ that you can use to find $dy/dt$.
A: First of all note that speed of man $(v_1)$ and speed of his shadow $(v_2)$ are constant values.Let us suppose that man starts walking from the bottom of the pole and let's observe case when man walks away another $2$ft from the point $C$ so that $CF=2$ft (see picture bellow). Note that triangle $BGD$ and triangle $BEH$ are similar so we may write next equation:
$GD : BD = EH : BE$
$2 : \sqrt{(AC)^2+(AB-CD)^2}=v_2(t_2-t_1): \sqrt{225+(v_2t_1)^2}$
where $t_1=\frac{AC}{v_1}$ ; $t_2=\frac{AC+2}{v_1}$ , so you have an equation with unknown $v_2$

