Posted by **Kara** on Monday, April 9, 2012 at 2:47pm.

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.

- Pre Cal -
**Henry**, Tuesday, April 10, 2012 at 10:27pm
.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.

- Pre Cal -
**Limy**, Wednesday, September 24, 2014 at 10:51pm
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

- Pre Cal -
**Kaylee**, Wednesday, April 27, 2016 at 8:18pm
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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