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Math Help - bearings

  1. #1
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    bearings

    An attendant in a lighthouse receives a request for aid from a stalled craft located 5 miles due east of the lighthouse. The attendant contacts a second boat located 14 miles from the lighthouse on a bearing of N23degreesW. What is the distance and the bearing of the rescue boat from the stalled craft. without using law of cosine


    sine23=x/14miles
    x=14xsine23
    x=5.47miles


    cos23=a/14miles
    a=14xcos23degrees
    a=12.89miles


    tan=sin/cos
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  2. #2
    MHF Contributor
    Joined
    Apr 2005
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    You're on your way to the soulution.

    So, as your intention,
    tan(theta) = a / (x+5)
    tan(theta) = 12.89 / (5.48 +5) = 1.23
    theta = arctan(1.23) = 50.89 degrees
    Or, maybe if you are looking for theta to the nearest drgree,
    theta = 51 degrees

    Hence, the bearing of the rescue boat from the stalled craft is
    N 39deg W -------------answer.

    The distance?

    By Pythagorean theorem,
    d = sqrt[(12.89)^2 +(5.48 +5)^2] = 16.61 miles. -----answer.

    By trig,
    sin(theta) = a / d
    d = 12.89 / sin(50.89deg) = 16.61 miles

    Or,
    cos(theta) = (x +5) / d
    d = (5.48 +5) / cos(50.89deg) = 16.61 miles.
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  3. #3
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    hello

    thanks for the help!!
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