A bird is sitting on the top of a vertical pole 20 m high and its elevation from a point O on the ground is `45o` . It flies off horizontally straight away from the point O. After one second, the elevation of the bird from O is reduced to `30o` . Then the speed (in m/s) of the bird is (1) `40(sqrt(2)-1)` (2) `40(sqrt(3)-2)` (3) `20""sqrt(2)` (4) `20(sqrt(3)-1)`

Explanation

in `/_ AOB`

`tan 45^@ = (AB)/(0B)`

`1= 20/(OB)`

`OB= 20m`

in`/_ COD`

`tan 30^@ = (CD)/(OD)`

`1/sqrt3 = 20/(20+x)`

`20 + x = 20 sqrt3`

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