A plane leaves JFK International Airport and travels due west at 570 mi/hr. Another plane leaves 20 minutes later and travels 22 degrees west of north at rate of 585 mi/hr. To the nearest ten miles, how far apart are they 40 minutes after the second plane leaves?
+130514 The first plane has traveled 570 mi
The second plane has traveled (2/3)hr * (585 mi/hr) = 390 miles
And the angle between them is 90 - 22 = 68°
And using the Law of Cosines, we have
d^2 = 570^2 + 390^2 - 2(570)(390)cos 68 ... where d is the distance between them after one hour
d^2 = 310449.9085672464
d = 557.18 miles = 560 miles {to the nearest ten miles}
+130514 The first plane has traveled 570 mi
The second plane has traveled (2/3)hr * (585 mi/hr) = 390 miles
And the angle between them is 90 - 22 = 68°
And using the Law of Cosines, we have
d^2 = 570^2 + 390^2 - 2(570)(390)cos 68 ... where d is the distance between them after one hour
d^2 = 310449.9085672464
d = 557.18 miles = 560 miles {to the nearest ten miles}
