related rates- rate a man's shadow changes as he walks past a lamp post (is a fixed distance away from it) A $186$ cm man walks past a light mounted $5$ m up on the wall of a building, walking at $2\ m/s$ parallel to the wall on a path that is $2$ m from the wall. At what rate is the length of his shadow changing when he is $4$ m past the point where the light is mounted? ($4$ m is the distance along the wall).
I have been doing related rates in my year 12 maths class and I know how to figure it out if the person is walking towards or away from the light but I've never come across a question like this where they are walking past the light, and have no idea where to begin. Some help getting started would be a massive help. (does it still involve similar triangles but in a 3D way?)

Let the distance along the wall of the man from the lamp be $x$ and let the length of the man's shadow, given $x$, be $\ell$.
Let $L$ be the point of the lamp, $P$ be the point at the bottom of the wall exactly under the lamp, $F$ be the point of the feet of the man, $H$ be the point of the top of the head of the man, and $E$ be the end of the man's shadow.

Consider that $EF=\ell$ and $FP=\sqrt{x^2+4}$. We draw the triangle $ELP$, including $F$ and $H$.

Here, by using similar triangles, $$\begin{align}&\frac{\ell}{1.86}=\frac{\ell+\sqrt{x^2+4}}{5}\\\to\ &5\ell=1.86\ell+1.86\sqrt{x^2+4}\\\to\ &3.14\ell=1.86\sqrt{x^2+4}\\\to\ &\ell=\frac{1.86}{3.14}\sqrt{x^2+4}\\\to\ &\ell=\frac{93}{157}\sqrt{x^2+4}\text.\end{align}$$
With the above equation, and considering that $\frac{dx}{dt}=v=2\ m/s$, we need to find $\frac{d\ell}{dt}$.
Consider that $$\ell^2=\frac{93^2}{157^2}x^2+\frac{4\cdot93^2}{157^2}\Rightarrow\frac{d\ell^2}{dx}=\frac{2\cdot93^2}{157^2}x\text.$$
It is clear that $$\frac{d\ell}{d\ell^2}=\frac1{2\ell}=\frac{157}{2\cdot93\sqrt{x^2+4}}\text.$$
By multiplying, we obtain $$\begin{align}\frac{d\ell}{dt}&=\frac{dx}{dt}\frac{d\ell^2}{dx}\frac{d\ell}{d\ell^2}\\&=2\cdot\frac{2\cdot93^2}{157^2}x\cdot\frac{157}{2\cdot93\sqrt{x^2+4}}\ m/s\\&=\frac{2\cdot93}{157}x\cdot\frac1{\sqrt{x^2+4}}\ m/s=\frac{186x}{157\sqrt{x^2+4}}\ m/s\text.\end{align}$$
Because we know that $x=4\ m$, we can substitute $4\ m$ for $x$:$$\begin{align}\frac{d\ell}{dt}&=\frac{186\cdot4}{157\sqrt{16+4}}\ m/s\\&=\frac{744}{314\sqrt5}\ m/s\\&=\frac{372\sqrt5}{157\cdot5}\ m/s\\&=\frac{372}{785}\sqrt5\ m/s\text.\end{align}$$
Therefore, the rate of change of the length of the man's shadow when the man is $4$ metres past the lamp is $\frac{372}{785}\sqrt5\ m/s$.
