An aeroplane flying horizontally at a height of 1.5km above the ground is observed at a certain points on earth to subtend an angle of 60. After 15 sec, its angle of elevation at the same point is observed to be 30. Calculate the speed of the plane in km/hr.
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[IN DIAGRAM, BC = ED = 1500 m. It is not 2500m. It is an error. I regret for the inconvenience]
In ∆ABC,
cot 60° = AC/BC
=> 1/√3 = AC/1500
=> AC = 1500/√3 ....(i)
In ∆AED,
cot 30° = AE/ED
=> √3 = AE/1500
=> AE = 1500√3 ....(ii)
From figure,
BD = CE
and, CE = AE - AC
so, BD = AE - AC
Now, substituting value of AE and AC from equations (i) and (ii),
BD = 1500√3 - 1500/√3
= 1500 (√3 - 1/√3)
= 1500 × 2/√3
BD = 3000/√3
BD is covered by the aeroplane in 15 seconds. So,
Speed of Aeroplane = Distance (BD) ÷ Time (t)
= 3000/√3 ÷ 15 m/sec
= 3000/ (15 × 1.732) m/sec
= 115.473 m/sec
Hence, speed of the aeroplane is 115.473
In ∆ABC,
cot 60° = AC/BC
=> 1/√3 = AC/1500
=> AC = 1500/√3 ....(i)
In ∆AED,
cot 30° = AE/ED
=> √3 = AE/1500
=> AE = 1500√3 ....(ii)
From figure,
BD = CE
and, CE = AE - AC
so, BD = AE - AC
Now, substituting value of AE and AC from equations (i) and (ii),
BD = 1500√3 - 1500/√3
= 1500 (√3 - 1/√3)
= 1500 × 2/√3
BD = 3000/√3
BD is covered by the aeroplane in 15 seconds. So,
Speed of Aeroplane = Distance (BD) ÷ Time (t)
= 3000/√3 ÷ 15 m/sec
= 3000/ (15 × 1.732) m/sec
= 115.473 m/sec
Hence, speed of the aeroplane is 115.473
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