The angle of elevation of an aeroplane from a point on the ground is 45° after flying for 15seconds, the elevation changes to 30° . If the aeroplane is flying at a height of 2500 meters, find the speed of the areoplane.

#### Solution

Let the height of flying of the aero-plane be PQ = BC and point A be the point of observation.

We have,

PQ = BC = 2500m, ∠PAQ = 45° and ∠BAC = 30°

In ΔPAQ,

` tan 45° = (PQ)/(AQ)`

`⇒ 1 = 2500/(AQ)`

`⇒AQ = 2500 m`

Also, in ΔABC,

`tan 30° = (BC)/(AC)`

`⇒1/ sqrt(3) = 2500/(AC)`

`⇒ AC = 2500 sqrt(3) m`

Now , QC = AC - AQ

`= 2500 sqrt(3)-2500`

`= 2500( sqrt(3)-1) m`

`= 2500 (1.732 - 1)`

`= 2500(0.732)`

`= 1830 m`

⇒ PB = QC = 1830

So, the speed of the aero-plane`(PB)/15`

`= 1830/15`

=122 m/s

`=122 xx3600/1000 km ⁄ h`

= 439.2 km /5

So, the speed of the aero-plane is 122m/ s or 439.2 km/ h.