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Finding Bearings

  1. Jun 8, 2013 #1
    Hey,
    I am studying for a test and i am having trouble with this problem. I have attached a photo of the worked solution. For the first part of the triangle (ABD) I understand everything except how they got the bearing AD. How do you find this bearing from the angle BAD.
     

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  2. jcsd
  3. Jun 9, 2013 #2

    haruspex

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    You know the angle ABD, and angle BAD has just been calculated. Angle ADC follows from these by a bit of basic geometry.
     
  4. Jun 9, 2013 #3
    Thanks for your help, but i'm still wondering how to find the bearing AD which is (208°00'20"). I found the angle for BDA (120°31'21") do i need to do something with that to find bearing AD, im still confused.
     
  5. Jun 9, 2013 #4

    haruspex

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    Right, so angle ADC is 180 minus that. You now have the bearing of DC and the angle ADC, so it's easy to deduce bearing DA. Just add/subtract 180 for AD.
     
  6. Jun 9, 2013 #5
    I looked over it again and that didn't really work out. So angle ADC is 180 minus 120°31'21" . This equals 59°28'39" you said this would equal the bearing DC, however the bearing DC is already given in the question which is 87°29'. I dont know what i did wrong.
     
  7. Jun 9, 2013 #6

    haruspex

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    I did not say that. I listed what you would know after you had worked out the angle ADC: the angle ADC and the bearing of DC. From those you can compute the bearing of DA.
     
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