An aeroplane leaves an in airport and flies due north at a speed of 100 km per hour at the same time another aeroplane leaves the same airport and flies due west at the speed of 1200 km per hour how far apart which with the two planes be after 1 and half hour
Answers
Distance travelled by aeroplane in 3/2 hours towards north direction = 1000*3/2
= 500*3
= 1500 km
So AB = 1500
Distance travelled by aeroplane in 3/2 hours towards north direction = 1200*3/2
= 600*3
= 1800 km
So BC = 1800
Now from Pythagorus theorem
AC2 = AB2 + BC2
=> AC2 = (1500)2 + (1800)2
=> AC2 = 2250000 + 3240000
=> AC2 = 5490000
=> AC = √5490000
=> AC =√(32 * 1002 * 61)
=> AC = 3*100* √61
=> AC = 300√61
So distance between two planes after 3/2 hours = 300√61 km
Answer:
Step-by-step explanation:
Distance travelled by aeroplane in 3/2 hours towards north direction = 1000*3/2
= 500*3
= 1500 km
So AB = 1500
Distance travelled by aeroplane in 3/2 hours towards north direction = 1200*3/2
= 600*3
= 1800 km
So BC = 1800
Now from Pythagorus theorem
AC2 = AB2 + BC2
=> AC2 = (1500)2 + (1800)2
=> AC2 = 2250000 + 3240000
=> AC2 = 5490000
=> AC = √5490000
=> AC =√(32 * 1002 * 61)
=> AC = 3*100* √61
=> AC = 300√61
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