The angle of elevation of a fighter from point A on the ground is `60^@` After a flight of 15 seconds the angle of elevation changes to `30^@`.If the jet is fighting at a speed of 720km/hour. Find the constant height at which the jet is flying.

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`.
From right `DeltaOMB`, we have
`(OM)/(BM) = cot 30^(2) = sqrt(3)`.
`rArr((x+3000))/(h) = sqrt(3) rArr x = (hsqrt(3) -3000)".........."(ii)`
Equating the values of x from (i) and (ii), we get
`h/(sqrt(3)) = h sqrt(3) - 3000 rArr h = 3h -3000sqrt(3)`.
`rArr 2h = 3000sqrt(3) rArr h = 500sqrt(3) = 1500xx 1.732 = 2598` .
Hence the required height is `2598 m`.