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Question
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.
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