An aeroplane flying horizontally at a height of 1.5 km above the
ground is observed at certain point on the earth to be subtend an angle of 60'. After 15 seconds, its angle of elevation is observed to be 30º. Calculate the speed of the
aeroplane in km/hr.
Answers
Answer:
[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 m/sec
❤️
Answer:
[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 m/sec
