The angle of elevation of a jet plane from a point A on the grund is `60^0` . After and flight of 30 seconds, the angle of elevation changes to `30^0dot` If the jet plane is flying at a constant height of `3600sqrt(3)m ,` find the speed of the jet plane.

Explanation

Let A be the point of observation and let AX be a horizontal line through A and `QC_|_AX`. Let P and Q bethe two position of the plane. Let `PB_|_AX`.

then , `PB = QC = 3600sqrt(3) m, angleBAP = 60^(@)` and `angleBAQ = 30^(@)`.
From right `DeltaABP`, we have
`(AB)/(BP) = cot 60^(@) = 1/(sqrt(3)) rArr (AB)/(3600sqrt(3)m) = 1/(sqrt(3))`
:Et `BC = PQ = x` metres.
Then, `AC = AB + BC = (x+3600) m` [using (i) ].
From right `DeltaACQ` , we have
`(AC)/(CQ) = cot 30^(@) = sqrt(3) rArr (x+3600)/(3600sqrt(3)) = sqrt(3)`
`rArr x + 3600 = (3600xx 3) = 10800`.
`rArr x = 10800 - 3600 = 7200`.
Thus , `PQ = 720`metres.
Now, `7200` m is covered in 30 seconds.
`:.`speed of the jet plane `= (7200/30 xx (60 xx 60)/(1000))km//hr`
`= 864km//hr`.
Hence, the speed of the jet plane is `864 km//hr`.