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`