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March 28, 2017

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A sail boat on lake Huron leaves Southampton and sails 20degrees west of north of 20km. At the same time, a fishing boat leaves Southampton and sails 30degrees west of south for 15 km. At this point, hot far apart are the boats, to the nearest metre? ( the answer is 4139m, but I need to know the method to understand how to get that answer)

  • Trig - ,

    distance is the difference in positions.

    But that involves vectors, so here, I think your teacher wants you to use trig.

    draw the figure.
    Notice the angles first: the interior angle between the paths is 130degrees

    Next, the law of cosines

    c^2=20^2+15^2-2*20*15Cos130
    that is it, solve for c.

  • Trig - ,

    Ok thank u:)

  • Trig - ,

    Um the answer that I got is 24.72km can u please help me get the answer above?

  • Trig - ,

    in my previous reply to your question,
    I made the silly mistake of adding 180-(20+15_ instead of 180-(20+30) to get the angle between their paths.
    http://www.jiskha.com/display.cgi?id=1279674199

    I agree with bobpursley's equation, and the result of that is
    c^2 = 400 + 225 - (-385.6726)
    c^2 = 1010.672566
    c = 31.8 km

    Your answer of 4139 m or 4.139 km makes no sense,
    I noticed you changed it from 4139km to 4139 m but it is still way off

  • Trig - ,

    Ok but do think u can help me with this question:
    a sailor out in a lake sees two likght houses 11km apart along the shore and gets bearings of 285degrees from his present position for light house A and 237degrees for light house B. From light house B, light house A has a bearing of 45degrees. How far to the nearest kilometre, is the sailor from each light house? What is the shortest distance, the nearest kilometre, from the sailor to the shore? ( the answers are 3km, 13km, 3km)

  • Trig - ,

    Using vector analysis, i calculated 31.79 km. So my cal. agree with bobpursley.

  • Trig - ,

    20km[110o] - 15km[240o] =
    -6.84+18.8i -(-7.5-13i) = 0.66 + 31.8i = 31.81 km[88.8o] N. of E.

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