A man of `2m` height walks at a uniform speed of `6kmh^(-1)` away from a lamp post of 6m height. The rate at which the length of his shadow increase in

Options

A. `2kmh^(-1)`
B. `1kmh^(-1)`
C. `3kmh^(-1)`
D. `6kmh^(-1)`

Answer

C

Explanation

In `DeltaADC, tan theta=6/(x+y)` and in `DeltasBCE, tan theta=2/x`
`:. 2/x=6/(x+y)impliesy=2x`

On differenting both sides w.r.t `t` we get
`(dy)/(dt)=2(dx)/(dt)implies(dx)/(dt)=3km//h [ :' (dy)/(dt)=6` given `]`