Answer :

Given that h_{m}=6ft, h_{l}=15ft, v_{m}=9ft/sec. We have to calculate v_{s} (speed of shadow).

After time t secs, the man would have moved distance 9t ft away from the lamp. Let the shadow move a distance of x ft in time t secs.

Consider ∆AEC and ∆BED

∠AED=∠BED=θ (common angle)

∠EAC=∠EBD=90°

Therefore, ∆AEC~∆BED by AA criteria

Substituting values, we get

Simplifying the equation, we get

x=6t – (1)

Differentiating (1) with respect to t, we get

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