The angle of the elevation of an aeroplane from a point on the ground is `45^(@)`. After flying for 15 seconds, the elevation changes to `30^(@)` . If the aeroplane is flying at a height of `2500` metres. Find the speed of the aeroplane.

Answer

`17.32 m , 30 m`

Explanation

Let A and B be the two positions of the aeroplane and let O be the point of observation.
From right `DeltaOCA`, we have
`(OC)/(AC) = cot 45^(2) rArr (OC)/(2500 m)= 1 rArr OC = 2500 m`.
From right `DeltaODB` , we have
`(OD)/(BD) = cot 30^(@) rArr (OD)/(2500 m) = sqrt(3) rArr OD = 2500sqrt(3) m`.
`:. AB = CD = OD- OC = 2500(sqrt(3) - 1) m = (25000 xx 0.732) m = 1830 m`.
Thus, the aeroplane covers `1830m` in `15` seconds.
`:.` Its speed `= ((1830)/(15) xx (60 xx 60)/(1000)) km//hr = 439.2 km//hr`.