Pre Cal

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.

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  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.

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  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

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  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.

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