A man who is $1.6\mathrm{ }\mathrm{m}$ tall walks away from a lamp which is $4\mathrm{ }\mathrm{m}$ above ground at the rate of $30\mathrm{ }\mathrm{m}/\min$. How fast is the man's shadow lengthening?

Let $PQ=4\mathrm{ }\mathrm{m}$ be the height of lamp and $AB=1.6\mathrm{ }\mathrm{m}$ be height of man. 

Let the end of shadow is $R$ and it is at a

distance of $l$ from $A$ when the man is at a distance of $x$

from $PQ$, at some instance.

Since, $∆PQR$ and $∆ABR$ are similar

$$\begin{gathered} \frac{PQ}{AB}=\frac{PR}{AR} \\ \Rightarrow \frac{4}{1.6}=\frac{x+l}{l} \end{gathered}$$

$$\begin{gathered} \Rightarrow 2x=3l \\ \Rightarrow 2\frac{dx}{dt}=3\frac{dl}{dt} \end{gathered}$$ given $\left[\frac{dx}{dt}=30m/min\right]$

$\therefore \frac{dl}{dt}=\frac{2}{3}·30\mathrm{ }\mathrm{m}/\min =20\mathrm{ }\mathrm{m}/\min$ (lengthening)

Hence, the required solution is $20m/min$.