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pallavi pradeep bhardwaj Grade: 12
        

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

7 years ago

Answers : (1)

Pratham Ashish
17 Points
										

4489-2361_4788_ques.jpg 


 


AB = height of lamp


CE = height of man


BC = y = distance b/w lamp n man


DC = x = length of shadow


Θ = angle ADB


now


tanΘ = 2/x (in smaller triangle)


tanΘ = 6/(x+y) .......[in larger triangle]


 


since both are same


so , 2/x = 6/(x+y)


=> 2x = y


x = y/2


now differentiating both sides w.r.to time ,t


dx/dt = (1/2)*dy/dt


dy/dt = 5m/hr.(given)


so, dx/dt = 2.5 m/hr.


 


 


 

7 years ago
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