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A ship sails 95 km on a bearing of 140 degree,then a further 102 km on a bearing of 260 degree and then returns directly to its starting point.find the length and bearing of the return journey.

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  1. I labeled my triangle ABC, with A as the starting point
    AB = 95, and BC = 102 , and angle ABC = 60°
    By the cosine law:
    AC^2 = 98^2 + 102^2 - 2(95)(102)cos60°

    find AC

    Once you have AC, use the sine law to find angle BAC and
    convert that into a bearing

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  2. why 98^2

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    2. 👎
  3. I labeled my triangle ABC, with A as the starting point
    AB = 95, and BC = 102 , and angle ABC = 60°
    By the cosine law:
    AC² = 95² + 102² - 2(95)(102)cos60
    = 98.686
    = 99 km

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