Question

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.

Answer 1

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