Answer :

Given that hm=6ft, hl=15ft, vm=9ft/sec. We have to calculate vs (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


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

The side of an eqMathematics - Board Papers

For the curve, <sMathematics - Board Papers

The money to be sMathematics - Board Papers

<span lang="EN-USMathematics - Board Papers

The side of an eqMathematics - Board Papers

The volume of a sMathematics - Board Papers