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

Options

A. 14.64m//s
B. 17.71m//s
C. 12m//s
D. None of these

Answer

A

Explanation

Let DB be the pole and the snake be at point O .
Then `angle BOD = 45^2` and BD = 20 meters .
Now the eagle flies horizontally and reaches at point M in 1 second .

Then `angle MON = 30 ^@ " where " MN _|_ ON`
Now , BD = MN = 20 meters
From triangle BOD,OD = 20 meters Again from `Delta MON`
` tan 30 ^@ = (MN)/(ON)=(20)/(20+DN)`
`therefore DN = 20 (sqrt(3)-1)= 20 xx 0.732 = 14.64` meters
`therefore " Speed of eagle "= "(Distance )"/"time"= BM/1=DN/1= 14.64 m//s`