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Question
An aeroplane is flying horizontally along a straight line at a height of 3000 m from the ground at a speed of 160 m/s. Find the time it would take for the angle of elevation of the plane as seen from a particular point on the ground to change from 60⁰ to 45⁰. Give your answer correct to the nearest second.
Solution
Given, AC = ED = 3,000 m
Speed = 160 m/s
In right ΔACB,
`tan 60^circ = "AC"/"BC"`
`=> sqrt3 = 3000/"BC"`
∴ BC = `3000/sqrt3`
`= 3000/sqrt3 xx sqrt3/sqrt3`
`= (3000 sqrt3)/3`
⇒ BC = `1000sqrt3` m
In right ΔEDB,
`tan 45^circ = "ED"/"BD"`
`=> 1 = 3000/"BD"`
⇒ BD = 3000 m
∴ AE = CD = BD - BC
∴ AE = 3000 - 1000`sqrt3`m
`= 1000 (3 - sqrt3)`
`= 1000 xx 1.268 ...(sqrt3 = 1.732)`
= 1268 m
∴ Time from A to E = `("Distance" ("AE"))/"Speed"`
`= 1268/160`
= 7.925 sec ≈ 8 sec
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