Laxmidevi standing on a railway overbridge of ` 5sqrt(3)` metres height observed the engine of the train from one side of the bridge of depression of `30^(@)` . But just after 2 seconds, she observed the engine at an angle of depression of `45^(@)` from the other side of the bridge. Find the speed of the train in metres per second.

Explanation

Let Laxmidevi standing at O observed the engine at first at A and then at B.
Also, let OC `bot` AB.
As per question,
OD = `5sqrt(3)` metres.
`angleEOA = angleOAD = 30^(@)`
and `angleBOF = angleOBD = 60^(@)`
Now, from the right-angled triangle AOD, we get,
`tan 30^(@) = (OD)/(AD)` [ by definition ]
or, `(1)/(sqrt(3)) = (5sqrt(3))/(AD) " " [ because " " OD = 5sqrt(3)m ]`
or, AD = 15.
Again, from the right-angled triangle BOD we get,
`tan 60^(@) = (OD)/(BD)`
or, `sqrt(3) = (5sqrt(3))/(BD) " " or, " "BD = 5`
So, the speed of the train = 20 metres per 2 seconds
=`(20)/(2)` metres per second
= 10 metres per second.
Hence the speed of the train = 10 m per second.