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Question
An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s a part is 30°, what is the speed of the aircraft?
Solution
The positions of the observer and the aircraft are shown in the given figure.
Height of the aircraft from ground, OR = 3400 m
Angle subtended between the positions, ∠POQ = 30°
Time = 10 s
In ΔPRO:
`tan 15^@ =(PR)/(OR)`
`"PR" = "OR" tan 15^@`
`=3400 xx tan 15^@`
`= 3400 xx 0.2679 = 910.86`
ΔPRO is similar to ΔRQO.
∴PR = RQ
PQ = PR + RQ
= 2PR
= 2 × 910.86 = 1821.72 m
∴ Speed of aircraft = `1821.72/10 = 182.2 "m/s"`
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