Pre Cal
posted by Kara .
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.

.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. 
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 
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.
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