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Mimi_x3

  • 4 years ago

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.

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  1. fiddlearound
    • 4 years ago
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    what do S30°W and 110° T mean ?

  2. Mimi_x3
    • 4 years ago
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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. fiddlearound
    • 4 years ago
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    |dw:1315630919336:dw|

  4. fiddlearound
    • 4 years ago
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    like that ^^^ ?

  5. fiddlearound
    • 4 years ago
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    |dw:1315631114054:dw|

  6. Mimi_x3
    • 4 years ago
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    Um, idk , it looks right though xDD

  7. Mimi_x3
    • 4 years ago
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    How can I find the distance between them ?

  8. fiddlearound
    • 4 years ago
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    |dw:1315631564224:dw|

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

  10. fiddlearound
    • 4 years ago
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    |dw:1315631903573:dw|

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

  12. fiddlearound
    • 4 years ago
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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.

  13. Mimi_x3
    • 4 years ago
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    Okay, tyvm (:

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

  15. Mimi_x3
    • 4 years ago
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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

  16. fiddlearound
    • 4 years ago
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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 :)

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

  18. fiddlearound
    • 4 years ago
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    what do you mean ?

  19. Mimi_x3
    • 4 years ago
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    |dw:1315632702671:dw|

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

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

  22. Mimi_x3
    • 4 years ago
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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 ?

  23. fiddlearound
    • 4 years ago
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    |dw:1315633004738:dw|

  24. Mimi_x3
    • 4 years ago
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    |dw:1315634036159:dw|

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

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

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

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

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

  30. upsilon
    • 4 years ago
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    Yes, that's it @fiddle

  31. Mimi_x3
    • 4 years ago
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    Yep that xD

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

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

  34. fiddlearound
    • 4 years ago
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    why did you pick 30 degrees ?

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

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

  37. fiddlearound
    • 4 years ago
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    |dw:1315634688115:dw|

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

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

  40. fiddlearound
    • 4 years ago
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    yeah, upsilon says the same thing.

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

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

  43. fiddlearound
    • 4 years ago
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    yep

  44. fiddlearound
    • 4 years ago
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    but the law of the cosines might work here.

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

  46. upsilon
    • 4 years ago
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    Yea, that will work

  47. fiddlearound
    • 4 years ago
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    |dw:1315635067683:dw|

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

  49. Mimi_x3
    • 4 years ago
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    Okay, tyvm (:

  50. fiddlearound
    • 4 years ago
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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 :)

  51. Mimi_x3
    • 4 years ago
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    Ok, ty (:

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

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

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