Question

A plane flies 320 km due west and then flies 240 km north. Find the shortest distance covered by the plane to reach its original position

Answer 1

Answer:

The first aeroplane leaves an airport and flies due north at a speed of 1000 km per hr.

∴ Distance travelled in 1.5 hrs.

=1000×1.5=1500 km

Similarly, Distance traveled by second aeroplane,

=1200×1.5=1800 km

In △ABC,

BC is distance travelled by first aeroplane and BA is the distance traveled by second aeroplane.

Hence, applying Pythagoras Theorem,

AC

2

=AB

2

+BC

2

∴ AC

2

=1500

2

+1800

2

∴ AC=300

61

km

Hence, two planes are 300

61

km apart

Answer 2

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