An airplane leaves an airport and flies due west 150 miles and then 230 miles in the direction S 39.67 degrees W. How far is the plane from the airport at this time to the nearest mile.

Answer 1

.id=150 Mi @ 180Deg. + 230 Mi @ 230.33Deg

X=150*cos180 + 230*cos230.33=-296.82 Mi
Y=150*sin180+230*sin230.33 = -177.04 Mi

d = sqrt((-296.82)^2+(-177.04)^2) = 346 Miles.

Answer 2

v = -(170 + 240 sin(69.5)) i - 240 cos(69.5) j .... is the resulting vector
|| v || = √((170 + 240 sin(69.5))² + (240 cos(69.5))²)
= 403.64892 mi

Answer: 404 mi

Answer 3

Limy is wrong; Henry is correct because im looking at this problem right now on my test review. The answer choices do not include 404.