Question

A plane files 320 km due west and then 2 to due north. Find the shortest distance covered by the plane to reach its original position.
Answer Plzzzz
fast​

Answer 1

• The total distance he has to travel to return to his starting point by the shortest route is 40 km.

• As the aeroplane flies due north and west the angle between west and north is 90°.

• The path travelled by the aeroplane starts from A and ends at C.

• The aeroplane travels 240 km north that is from A to B and 320 km west that is from B to C.

• Now we have to find the distance between A and C to compute the shortest route to return to its starting point.

Using Pythagoras Theorem,

AC² = AB² + BC²

AC² = 240² + 320²

AC² = 5760 + 10240

AC² = 160000

AC = 400 km

The distance aeroplane needs to travel by the shortest distance is 400 km.

Answer 2

Answer:

320km west and 2km north.

shortest distance can be found using Pythagoras theorem.

Here a²+b²=c²

applying formula

(320)²+(2)²=c²

c²=102400+4

c²=102404

c=√102404

c=320 km

due north west

so the shortest distance would approximately be 320 km due North-West.

thank you