A man of height 1.7 m walks at a uniform speed of 6.6 m/min from a lamp post which is 5m high.Find the rate at which the length of his shadow increases.

Explanation

Let AB is a lamp post of height 5 m. Let at any time 't', a man of height 1. 7 m is at a distance 'y' metre from the lamp post and its shadow is CE = x metre.
Let `angleCED=theta`
`In DeltaCED,`
`tan theta=(CD)/(CE)`

And in `DeltaABE,`
`tan theta = (AB)/(AE)`
`:." "(CD)/(CE)=(AB)/(AE)`
`rArr " "(1.7)/x=5/(x+y)`
`rArr " "5x=1.7x+1.7y`
`rArr " "3.3x=1.7y`
`rArr " "x=(17)/(33) y`
`:. " "(dx)/(dt)=(17)/(33)xx(dy)/(dt)`
`" "=(17)/(33) xx6.6=3.4 m//min.`
Therefore the shadow is increasing at the rate of 3.4 m/min.