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
