A plane leaves jfk international airport and travels due west at 570 mi/hr. another plane leaves 20 minutes later and travels 22º west of north at the rate of 585 mi/h. to the nearest ten miles, how far apart are they 40 minutes after the second plane leaves.

Distance=speed*time

distance traveled by plane A after 40 minutes:

distance=570*40/60=380 miles

Distance traveled by B after 40 minutes

distance=20/60*585=195 miles

Thus the distance between them will be given using cosine law:

c^2=a^2+b^2-2ab Cos C

thus we shall have:

c^2=380^2+195^2-2*380*195*cos22

c^2=45016.35275

c=212.171 miles

Thus the distance between them after 40 minutes is 212.171 miles