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BloomLocke367

  • one year ago

http://prntscr.com/7bvs3w @Nnesha

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  1. Nnesha
    • one year ago
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    draw a triangle and find 3rd side

  2. BloomLocke367
    • one year ago
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    what do you mean? (sorry, I really don't understand trig)

  3. Nnesha
    • one year ago
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    |dw:1433170800461:dw| given order pair (x,y) x represent cosine y represent sin and \[\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }\]

  4. BloomLocke367
    • one year ago
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    how do I find csc and sec?

  5. Nnesha
    • one year ago
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    first you have to find 3rd side of right triangle |dw:1433171058392:dw| sin = opposite /hyp and cos = adjacent /hyp

  6. Nnesha
    • one year ago
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    now apply Pythagorean theorem to find 3rd side

  7. BloomLocke367
    • one year ago
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    13

  8. Nnesha
    • one year ago
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    yep right |dw:1433171191619:dw| now what is reciprocal of sec and csc ?? for example tan = sin/cos sec= ?

  9. BloomLocke367
    • one year ago
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    I don't know

  10. Nnesha
    • one year ago
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    mhm that's important part of trig you have to know this \[\huge\rm sec = \frac{ 1 }{ \cos } , \csc =\frac{ 1 }{ \sin } ,\cot =\frac{ \cos }{ \sin } \]

  11. Nnesha
    • one year ago
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    cos is adjacent over hyp so 1/adjacent/hpy = ? what is adjacent of right triangle

  12. BloomLocke367
    • one year ago
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    12

  13. BloomLocke367
    • one year ago
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    so 1/12/13?

  14. Nnesha
    • one year ago
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    yep right now multiply top wit the reciprocal of the bottom

  15. BloomLocke367
    • one year ago
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    hang on, I have to volunteer now, I'll be back in 2 hours, approximately.

  16. Nnesha
    • one year ago
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    aright :-)

  17. BloomLocke367
    • one year ago
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    @Nnesha I'm back.. I know it took forever, but I never got the chance after I finished working

  18. BloomLocke367
    • one year ago
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    I got 1? is that right? @Nnesha

  19. Nnesha
    • one year ago
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    nope how did you get 1?

  20. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @BloomLocke367 so 1/12/13? \(\color{blue}{\text{End of Quote}}\) multiply top wit the reciprocal of the bottom fraction

  21. BloomLocke367
    • one year ago
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    ohh I didn't use the reciprocal

  22. BloomLocke367
    • one year ago
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    I got 169/144

  23. Nnesha
    • one year ago
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    mhm how did you get 169 and 133 don't square them |dw:1433196171493:dw|

  24. BloomLocke367
    • one year ago
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    what?

  25. Nnesha
    • one year ago
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    5,12 aand 13 are sides of right triangle what is cos = ? sin = ?

  26. Nnesha
    • one year ago
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    \[\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }\]

  27. BloomLocke367
    • one year ago
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    cos is 12/13 and sin is 5/13

  28. Nnesha
    • one year ago
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    yes right now you need sec and csc so sec= 1/cos which is \[\frac{ 1 }{ \frac{12}{13}}\]

  29. Nnesha
    • one year ago
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    multiply 1 by the reciprocal of the bottom fraction

  30. Nnesha
    • one year ago
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    for example \[\huge\rm \frac{ \frac{ a }{ b } }{ \frac{ c }{ d } } = \frac{ a }{ b } \times \frac{ d }{ c }\]

  31. BloomLocke367
    • one year ago
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    so 13/12

  32. Nnesha
    • one year ago
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    yeah!

  33. Nnesha
    • one year ago
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    csc = ?

  34. BloomLocke367
    • one year ago
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    and that's sec? that's not an option...

  35. Nnesha
    • one year ago
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    i did it wrong |dw:1433196991226:dw| sin is opposite so (cosine , sin) x coordinate would be adjacent which suppose to be 5

  36. Nnesha
    • one year ago
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    so cos = 5/13 sin = 12/13

  37. Nnesha
    • one year ago
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    so csc = what ? and sec = ?

  38. BloomLocke367
    • one year ago
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    sec=13/5 and csc=13/12

  39. BloomLocke367
    • one year ago
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    Thank you so much!

  40. Nnesha
    • one year ago
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    np :-)

  41. BloomLocke367
    • one year ago
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    Can you help more?

  42. Nnesha
    • one year ago
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    mhm maybe...

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