Related rates shadow question A $5$ meter lamp is casting a shadow on a $1.8$ meter man walking away at $1.2$ meters a second, how fast is the shadow increasing?
I have no idea how to do this, it feels like there is missing information. I know that this is a problem about triangles but there is some weird trick that has to be used since only two heights are known which are both the same part of a triangle.
 A: After $t$ seconds, the man has traveled $1.2t$ meters from the lamp. Let $\mathrm{shadow}(t)$ denote the length of the man's shadow after $t$ seconds.

Note that triangles $\triangle ABC$ and $\triangle CDE$ are similar. Therefore the ratios between corresponding sides must be equal:
$$\frac{BC}{AB}=\frac{DE}{CD}$$
which tells us
$$\frac{1.2t}{3.2}=\frac{\mathrm{shadow}(t)}{1.8}$$
so that
$$\mathrm{shadow}(t)=\frac{1.2t}{3.2}\times 1.8=\frac{2.16 t}{3.2}=\frac{27t}{40}.$$
Therefore
$$\frac{d\,\mathrm{shadow}(t)}{dt}=\frac{d}{dt}\left(\frac{27t}{40}\right)=\frac{27}{40}\,\text{m/s}$$
A: Let $x$ be the horizontal distance between the lamp and the man and let $s$ be the length of the shadow. Note that $\dfrac{dx}{dt} = 1.2$ and we are asked to find $\dfrac{ds}{dt}$. By similar triangles (draw a diagram), we have:
$$
\dfrac{1.8}{5} = \dfrac{s}{s+x} \iff 1.8s+1.8x=5s \iff 3.2s=1.8x \iff s = \dfrac{9}{16}x
$$
Hence, by taking the derivative of both sides with respect to $t$, we obtain:
$$
\dfrac{ds}{dt} = \dfrac{9}{16}\dfrac{dx}{dt} = \dfrac{9}{16}(1.2) = \dfrac{27}{40} \text{m/s}$$
A: I don't think that there's any missing information here.  At any given time $t$, we can construct two similar triangles as follows:
Let $A$ be the top of the man, let $B$ be at the man's feet, and let $C$ be the edge of his shadow.  Let $A'$ be the top of the lamppost and let $B'$ be the bottom of the lamppost.  We note that the triangles $ABC$ and $A'B'C$ are similar.  
Now, let $s(t)$ be the length of the man's shadow, and let $x(t)$ be the distance of the man from the lamppost.  By the above, we have the proportion
$$
\frac{5\,m}{1.8\,m}=\frac{x(t)+s(t)}{s(t)} = 1+\frac{x(t)}{s(t)}
$$
That should be enough to let you deduce the nature of $s(t)$, the movement of the shadow.
