Monu standing in the midst of a field, observes a flying bird in his north at an angle of elevation of `30^(@)` and after 2 minutes he observer the bird in his south at an angle of elevation of `60^(@)`. If the bird flies in a straight line all along at a height of `50sqrt(3)` metres, find its speed in kilometre per hour.

Explanation

Let Monu standing at O observes the flying bird in his north at the point N at an angle of elevation of `30^(@)` and in his south at the point S at an angle of elevation of `60^(@)`.
Also, let OC `bot` NS and AB is the horizontal line.
As per question, OC = `50sqrt(3)` metres.
`angleAON = angleONC = 30^(@) and angleBOS = angleOSC = 60^(@)`
Now, from the right-angled triangle CON we get,
` tan30^(@) = (OC)/(CN)` [ by definiton]
or, `(1)/(sqrt(3)) = (50sqrt(3))/(CN) " " or, " " CN = 150`
Again, from the right-angled triangle COS, we get,
`tan 60^(@) = (OC)/(CS)` [by definition ]
or, `sqrt(3) = (OC)/(CS) " " or, " "sqrt(3) = (50sqrt(3))/(CS) " "` or, CS = 50.
`therefore " " `NS = CN + CS
=150 + 50
= 200
So, the speed of the bird = 200 metres per 2 minutes
= `(200)/(2)` metres per 1 "
= `(200xx60)/(2xx1000)` km/h
= 6 km per hour
Hence the speed of the bird = 6 km per hour.