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anonymous

  • one year ago

I got an answer for this trig problem and I want to see if it is correct

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    I know the answer is either approximately 50.3m or approximately 39.3m, but I do not know which one it is

  3. anonymous
    • one year ago
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    @jim_thompson5910

  4. IrishBoy123
    • one year ago
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    did you draw the picture? rough sketch will suffice

  5. anonymous
    • one year ago
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    i did

  6. IrishBoy123
    • one year ago
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    |dw:1438642173189:dw|

  7. anonymous
    • one year ago
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    |dw:1438642164202:dw|

  8. anonymous
    • one year ago
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    Here is what I did

  9. anonymous
    • one year ago
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    @mathstudent55

  10. anonymous
    • one year ago
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    @IrishBoy123 did I solve the problem correctly

  11. IrishBoy123
    • one year ago
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    why did you not use cos 43 ?

  12. IrishBoy123
    • one year ago
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    https://www.mathsisfun.com/algebra/trig-cosine-law.html

  13. anonymous
    • one year ago
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    wait why would it be cos 43

  14. IrishBoy123
    • one year ago
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    well look at the link just asking

  15. anonymous
    • one year ago
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    I guess that it is right I think

  16. amistre64
    • one year ago
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    law of cosine is fine, but i dont know how to read the bearings ...

  17. IrishBoy123
    • one year ago
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    |dw:1438642650401:dw|

  18. anonymous
    • one year ago
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    So my new equation is \[d=\sqrt{30^2+34^2-2(30)(24)(\cos(43}))\]

  19. IrishBoy123
    • one year ago
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    yes but that has to make sense to you.... again http://mathworld.wolfram.com/LawofCosines.html

  20. IrishBoy123
    • one year ago
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    and typo (!) : \(34^2 -> 24^2\)

  21. anonymous
    • one year ago
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    Your graph made better sense than the link, thank you though, therefore my new anser should be, 20.6m

  22. IrishBoy123
    • one year ago
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    if **you** are happy with it in the attached diagram we can say - from Law of Cosines - that \( \large a^2 = b^2 + c^2 - 2 b c\ cos A\) |dw:1438643019135:dw|

  23. anonymous
    • one year ago
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    ok so is my new anser 20.6m

  24. IrishBoy123
    • one year ago
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    |dw:1438643119009:dw| and \(b^2 = a^2 + c^2 - 2ac \ cos B\)

  25. IrishBoy123
    • one year ago
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    it is Pythagoreas expanded |dw:1438643242132:dw| so \(\large c^2 = a^2 + b^2 - 2ab \ cos C\)

  26. anonymous
    • one year ago
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    ok, but can you confirm that I have the right answer

  27. IrishBoy123
    • one year ago
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    and yes i seem to agree with you

  28. anonymous
    • one year ago
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    ok so the answer to my original problem is20.6

  29. IrishBoy123
    • one year ago
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    i should think so

  30. IrishBoy123
    • one year ago
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    |dw:1438643502160:dw| look at this, where C = 90 \(c^2 = a^2 + b^2 - 2ab \ cos C = a^2 + b^2 - 2ab \ (0)\) \(c^2 = a^2 + b^2\)

  31. anonymous
    • one year ago
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    Can you help me with a simpler problem that I also dont know whether I got right or wrong

  32. anonymous
    • one year ago
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    @jdoe0001

  33. jdoe0001
    • one year ago
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    repost anew, thus more eyes :)

  34. jdoe0001
    • one year ago
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    20.6 is correct btw

  35. anonymous
    • one year ago
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    ok

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