SOLUTION: Two planes leave an airport at the same time, one flying due west at 500km/h and the other flying due southeast at 300km/h. What is the distance between the planes two hours later?

Algebra ->  Trigonometry-basics -> SOLUTION: Two planes leave an airport at the same time, one flying due west at 500km/h and the other flying due southeast at 300km/h. What is the distance between the planes two hours later?      Log On


   



Question 43142: Two planes leave an airport at the same time, one flying due west at 500km/h and the other flying due southeast at 300km/h. What is the distance between the planes two hours later?
Answer by psbhowmick(878) About Me  (Show Source):
You can put this solution on YOUR website!
Note: Distance = Speed x Time

In 2 hours, the plane travelling to West travels a distance of 500x2 = 1000 km.
In 2 hours, the plane travelling to South-East travels a distance of 300x2 = 600 km.

If their initial position was the point O then their final positions, after 2 hours, are points A and B respectively.


We need to know the length of AB.
In triangle BCO, as OB is in South-East direction.
So, OC%2FOB = cos(< BOC) = cos%2845%5Eo%29
or OC%2F600+=+cos%2845%5Eo%29
or OC%2F600+=+1%2Fsqrt%282%29
or OC+=+600%2Fsqrt%282%29
or OC+=+300sqrt%282%29
or OC = 424.264 km

Therefore, AC = AO + OC = 1000 + 424.264 = 1424.264 km

As BCO is a right angled isosceles triangle with
Now, in triangle ACB, Applying Pythagorus theorem,
AB%5E2+=+AC%5E2+%2B+BC%5E2
or AB%5E2+=+1424.264%5E2+%2B+424.264%5E2
or AB%5E2+=+2208528.14
or AB+=+sqrt%282208528.14%29
or AB = 1486.112 km

Hence, the reqd. distance between the two planes after 2 hours is 1486.112 km.