A bird is sitting on the top of a vertical pole 20 m high and its elevation fron a point O on the ground is `45^@`,It Files off horizontally straight away from the point O.After one second , the elevation of the bird fron O is reduced to `30^@` Then the speed (in m/s) of the bird is

Options

A. `20sqrt2`
B. `20(sqrt3-1)`
C. `40(sqrt2-1)`
D. `40(sqrt3-sqrt2)`

Answer

B

Explanation

Let AB be the vertical pole of height 20 m and let Q be the position of bird after one second . It is given that `angleAOB=45^@` and `anglePOQ=30^@`

In `triangleOAB` and OPQ, we have
`tan 45^@ =(OB)/(OA)` and `tan 30^@=(PQ)/(OP)`
`rArr OA=20` and `sqrt3PQ=OP`
`rArr OP=20sqrt3 " " [because PQ=OA=20]`
`rArr OA+AP = 20sqrt3`
`rArr AP=20sqrt3-20=20(sqrt3-1)`