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Mimi_x3 Group Title

Amy is 20m S30°W from a tree. Belinda is 50m 110° T from the same tree. Find the distance between them, to the nearest metre.

  • 3 years ago
  • 3 years ago

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  1. fiddlearound Group Title
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    what do S30°W and 110° T mean ?

    • 3 years ago
  2. Mimi_x3 Group Title
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    Um, I'm not sure but i think South 30 degrees West , 110 degrees true bearing idk. I haven't done it in ages so I forgot xD

    • 3 years ago
  3. fiddlearound Group Title
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    |dw:1315630919336:dw|

    • 3 years ago
  4. fiddlearound Group Title
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    like that ^^^ ?

    • 3 years ago
  5. fiddlearound Group Title
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    |dw:1315631114054:dw|

    • 3 years ago
  6. Mimi_x3 Group Title
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    Um, idk , it looks right though xDD

    • 3 years ago
  7. Mimi_x3 Group Title
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    How can I find the distance between them ?

    • 3 years ago
  8. fiddlearound Group Title
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    |dw:1315631564224:dw|

    • 3 years ago
  9. Mimi_x3 Group Title
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    Um, I don't think that it requires sin and cos, its bearings

    • 3 years ago
  10. fiddlearound Group Title
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    |dw:1315631903573:dw|

    • 3 years ago
  11. Mimi_x3 Group Title
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    What? There's not values of x and y, how can you use the distance formula ?

    • 3 years ago
  12. fiddlearound Group Title
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    I got the values of x and y using the sin() and cos() in the table I drew above. Rounded to the nearest metre, I get 64 meters distance.

    • 3 years ago
  13. Mimi_x3 Group Title
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    Okay, tyvm (:

    • 3 years ago
  14. fiddlearound Group Title
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    does it make sense ? did you learn this stuff in the past ?

    • 3 years ago
  15. Mimi_x3 Group Title
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    Not in the past xD, I learnt it year 7 and now I forgot xD But I remember that there's an easier way that doesn't require sin and cos

    • 3 years ago
  16. fiddlearound Group Title
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    Well , if there is an easier way, I hope someone will post it here - coz I'd like to know it too :)

    • 3 years ago
  17. Mimi_x3 Group Title
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    Hey, you know your diagram what is the angle degrees thing for AB

    • 3 years ago
  18. fiddlearound Group Title
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    what do you mean ?

    • 3 years ago
  19. Mimi_x3 Group Title
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    |dw:1315632702671:dw|

    • 3 years ago
  20. fiddlearound Group Title
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    yes we can calculate that - it's 240-110 = 130 degrees

    • 3 years ago
  21. fiddlearound Group Title
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    and you're right that we can make the problem simpler

    • 3 years ago
  22. Mimi_x3 Group Title
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    Yep, there's a formula to calculate it then, was it the sine or cosine formula ? or was it soh cah toa ?

    • 3 years ago
  23. fiddlearound Group Title
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    |dw:1315633004738:dw|

    • 3 years ago
  24. Mimi_x3 Group Title
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    |dw:1315634036159:dw|

    • 3 years ago
  25. fiddlearound Group Title
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    I would need to calculate to find out - why are you asking ?

    • 3 years ago
  26. Mimi_x3 Group Title
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    If that angle can be found, then I can use the sine rule xD

    • 3 years ago
  27. fiddlearound Group Title
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    are you trying to use the law of the cosines ?

    • 3 years ago
  28. Mimi_x3 Group Title
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    No, the Sine Rule it says it in my book xD

    • 3 years ago
  29. fiddlearound Group Title
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    which sine rule ? a/sin(A)=b/sin(B)=c/sin(C) ?

    • 3 years ago
  30. upsilon Group Title
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    Yes, that's it @fiddle

    • 3 years ago
  31. Mimi_x3 Group Title
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    Yep that xD

    • 3 years ago
  32. fiddlearound Group Title
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    yes, but we only have one angle , that 130 degrees at this point.

    • 3 years ago
  33. Mimi_x3 Group Title
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    I know , but the other angle is 30 right ? the one that I drew

    • 3 years ago
  34. fiddlearound Group Title
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    why did you pick 30 degrees ?

    • 3 years ago
  35. fiddlearound Group Title
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    is it becasue sin(30)=1/2 ?

    • 3 years ago
  36. Mimi_x3 Group Title
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    Well, the question S30°W so that angle is 30 right ?

    • 3 years ago
  37. fiddlearound Group Title
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    |dw:1315634688115:dw|

    • 3 years ago
  38. upsilon Group Title
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    No, you can't say that...@mimi|dw:1315634688202:dw|

    • 3 years ago
  39. fiddlearound Group Title
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    The 30 degrees would be relative to the black line I drew.

    • 3 years ago
  40. fiddlearound Group Title
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    yeah, upsilon says the same thing.

    • 3 years ago
  41. Mimi_x3 Group Title
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    Can it be 60 then, since 90-30 = 60 ?

    • 3 years ago
  42. upsilon Group Title
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    You can't decide like that you need the angle to be in ratio with the opposite sides...

    • 3 years ago
  43. fiddlearound Group Title
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    yep

    • 3 years ago
  44. fiddlearound Group Title
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    but the law of the cosines might work here.

    • 3 years ago
  45. fiddlearound Group Title
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    http://hyperphysics.phy-astr.gsu.edu/hbase/lcos.html

    • 3 years ago
  46. upsilon Group Title
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    Yea, that will work

    • 3 years ago
  47. fiddlearound Group Title
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    |dw:1315635067683:dw|

    • 3 years ago
  48. upsilon Group Title
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    \[d = \sqrt{ a^2 + b^2 - 2*a*b*cos\alpha}\]

    • 3 years ago
  49. Mimi_x3 Group Title
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    Okay, tyvm (:

    • 3 years ago
  50. fiddlearound Group Title
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    http://www.wolframalpha.com/input/?i=sqrt%2820^2%2B50^2-2*20*50*cos%28130%29%29 It's 64 , same answer I got with the first method :)

    • 3 years ago
  51. Mimi_x3 Group Title
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    Ok, ty (:

    • 3 years ago
  52. fiddlearound Group Title
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    and it's much shorter than the first method :)

    • 3 years ago
  53. Mimi_x3 Group Title
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    yup xD. Your method looked so complicated before (:

    • 3 years ago
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