Question

at the same rate?
A girl of height 1.6 m walks at the rate of 50 metres per minute away from a lamp whis
is 4 m above the ground. How fast is the girl's shadow lengthening?

​

Answer 1

Answer:

33.3 m / min

Step-by-step explanation:

Given  

at the same rate? A girl of height 1.6 m walks at the rate of 50 metres per minute away from a lamp which 4 m above the ground.  

In Δ RST

 Tan θ = RS / SE

           = 1.6 / b -------------1

In Δ PQT

Tan θ = PQ / QT

         = 4 / a + b ------------2

From 1 and 2 we get

1.6 / b = 4 / a + b  

0.4 a + 0.4 b = b

0.4 a = (1 – 0.4) b

0.4 a = 0.6 b

2 a = 3 b

Da/dt = 50 m / min

Now shadow length increases as he walks.

2 da / dt = 3 db /dt

So db / dt = 2/3(da / dt)

Db / dt = 2/3 x 50  

            = 100/3  

             = 33.3 m / min  

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