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Question

A man of height 2m walks at uniform speed of 5km/hr away from a lamp post which is 6m high. Find the rate of which the length of his shadow increases.

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Solution

Let AB be the lamp post
and let MN be the man of height 2m
and let AM=lmeter.
and MS is the shadow of the man.
Let length of shadow MS=s
Given: Man walks at speed of 5kmph
dldt=5kmph
We need to find rate at which the length of his shadow increases=dsdt.

In ASB,tanθ=ABAS=6l+s ....(1)

In MSN,tanθ=MNMS=2s ....(2)

From (1) and (2)

6l+s=2s

6s=2l+2s

4s=2l

or l=2s

dldt=2dsdt

5=2dsdt since given that dldt=5kmph

dsdt=52kmph=2.5kmph

1265141_1373109_ans_27e585004f8b421e8d6338de7e7da58c.PNG

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