An aeroplane leaves an airport and flies due west at a speed of 2100 km/hr. At the same time, another aeroplane leaves the same place at airport and flies due south at a speed of 2000 km/hr. How far apart will be the two planes after 1 hour?
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Let O be the position of an airport.
In 1 hour the distance travelled by aeroplane when it flies due West at a speed of 2100 km/h
OA = 2100 × 1 = 2100 km
[Distance = speed × time]
In 1 hour the distance travelled by aeroplane when it flies due South at a speed of 2100 km/h
OA = 2000 × 1 = 2000 km
[Distance = speed × time]
In ∆ AOB
AB² = OB² + OA²
[By Pythagoras theorem]
AB² =( 2100)² + (2000)²
AB² = 4410000 + 4000000
AB² = 8410000 km
AB = √ 8410000
AB √ 2900 × 2900
AB = 2900 km
Hence, the distance between two planes after one hour is 2900 km.
HOPE THIS WILL HELP YOU..
In 1 hour the distance travelled by aeroplane when it flies due West at a speed of 2100 km/h
OA = 2100 × 1 = 2100 km
[Distance = speed × time]
In 1 hour the distance travelled by aeroplane when it flies due South at a speed of 2100 km/h
OA = 2000 × 1 = 2000 km
[Distance = speed × time]
In ∆ AOB
AB² = OB² + OA²
[By Pythagoras theorem]
AB² =( 2100)² + (2000)²
AB² = 4410000 + 4000000
AB² = 8410000 km
AB = √ 8410000
AB √ 2900 × 2900
AB = 2900 km
Hence, the distance between two planes after one hour is 2900 km.
HOPE THIS WILL HELP YOU..
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Answer:
Step-by-step explanation:
Let O be the position of an airport.
In 1 hour the distance travelled by aeroplane when it flies due West at a speed of 2100 km/h
OA = 2100 × 1 = 2100 km
[Distance = speed × time]
In 1 hour the distance travelled by aeroplane when it flies due South at a speed of 2100 km/h
OA = 2000 × 1 = 2000 km
[Distance = speed × time]
In ∆ AOB
AB² = OB² + OA²
[By Pythagoras theorem]
AB² =( 2100)² + (2000)²
AB² = 4410000 + 4000000
AB² = 8410000 km
AB = √ 8410000
AB √ 2900 × 2900
AB = 2900 km
Hence, the distance between two planes after one hour is 2900 km.
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