$\mathrm{A}$ 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?

It is given that a man who is $1.6$ m tall is walking away from a lamp which is $4$m above the ground at the rate of $30 m/min$.

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

Let at any instant, the end of the shadow be denoted by $R$  which is at a distance of $\text{'}l\text{'}$ m from $A$ when the man is at a distance of $x$ m from $PQ$.

The following setup can be taken for reference.

Now, consider the two triangles $ARB$and $PRQ$.

Here, $\angle QPR = \angle BAR = 90°$ and $\angle PRQ = \angle ARB (Common angle)$

As two angles of one triangle are equal to two of the other, we can say that the triangles are similar.

As we know that the ratio of corresponding sides of two similar triangles are equal, we get,

$$\begin{gathered} \frac{AR}{PR}=\frac{AB}{PQ} \\ \Rightarrow \frac{l}{l+x}= \frac{1.6}{4}=\frac{2}{5} \\ 5l = 2l + 2x \\ l=\frac{2}{3}x \end{gathered}$$

On differentiating both sides w.r.t $t$, we get,

$\frac{dl}{dt}=\frac{2}{3}\times \frac{dx}{dt}$

We know that rate of increase of length of the shadow is given by $\frac{dl}{dt}$ while $\frac{dx}{dt}$ is the velocity with which the man moves away from the lamp.

$\Rightarrow$ Rate of increase of length of shadow $= \frac{2}{3}\times Velocity of man = \frac{2}{3}\times 30 = 20m/min$.

Hence, our final answer is $20 m/min$.