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(B) Let $AB$ be the lamp post of height $4.5 \ m$ and $PQ$ be the man of height $1.8 \ m$. Let the man be at a distance $y$ from the lamp post,so $QB

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$0.8$

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(B) Let $AB$ be the lamp post of height $4.5 \ m$ and $PQ$ be the man of height $1.8 \ m$. Let the man be at a distance $y$ from the lamp post,so $QB = y$. Let the length of the shadow be $x$,so $CQ = x$.From similar triangles $	riangle AB C$ and $	riangle PQC$,we have:$\frac{PQ}{AB} = \frac{CQ}{CB}$$\frac{1.8}{4.5} = \frac{x}{x + y}$$\frac{2}{5} = \frac{x}{x + y}$$2(x + y) = 5x$$2x + 2y = 5x$$3x = 2y$$x = \frac{2}{3}y$Differentiating with respect to time $t$:$\frac{dx}{dt} = \frac{2}{3} \frac{dy}{dt}$Given that the man is moving away at $\frac{dy}{dt} = 1.2 \ m/sec$,we have:$\frac{dx}{dt} = \frac{2}{3} 	imes 1.2 = 0.8 \ m/sec$.Thus,the shadow is lengthening at a rate of $0.8 \ m/sec$.

$A$ man of height $1.8 \ m$ is moving away from a lamp post at the rate of $1.2 \ m/sec$. If the height of the lamp post is $4.5 \ m$,then the rate at which the shadow of the man is lengthening is .......... $m/sec$.

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