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`        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`
8 years ago

Pratham Ashish
17 Points
```

AB = height of lamp
CE = height of man
BC = y = distance b/w lamp n man
DC = x = length of shadow
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.

```
8 years ago
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