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anonymous

  • one year ago

Explain SOH-CAH-TOA sin=______ cos=______ tan=_______

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  1. anonymous
    • one year ago
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    @peachpi

  2. triciaal
    • one year ago
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    |dw:1440640439967:dw|

  3. triciaal
    • one year ago
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    |dw:1440640548365:dw|

  4. anonymous
    • one year ago
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    O thanks :)

  5. triciaal
    • one year ago
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    SOH -CAH-TOA is a "code" used to remember the trig identities this is with a right triangle the hypotenuse is always opposite the right angle and is the longest side of the triangle

  6. anonymous
    • one year ago
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    If i wanted to know sin cos and tan but in terms of not using opp hyp and adj. what would they look like?

  7. anonymous
    • one year ago
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    @triciaal

  8. anonymous
    • one year ago
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    they would be points on the unit circle where the input was the angle, and the output for cosine is the first coordinate and for sine the second coordinate of that point

  9. anonymous
    • one year ago
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  10. anonymous
    • one year ago
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    i kind of dont understand

  11. anonymous
    • one year ago
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    @satellite73

  12. anonymous
    • one year ago
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    if you are using the mnemonic device soh cah toa , save what i sent for later it will be a big help believe me, but you are not ready for it yet

  13. anonymous
    • one year ago
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    ok

  14. anonymous
    • one year ago
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    i just dont know what to put in the blanks in number 5

  15. anonymous
    • one year ago
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    oh i think they want you to say\[\sin(\theta)=\frac{y}{r},\cos(\theta)=\frac{x}{r},\tan(\theta)=\frac{y}{x}\]

  16. anonymous
    • one year ago
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    whoever wrote this made a very serious error however

  17. anonymous
    • one year ago
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    what was the error? It was my teacher btw lol

  18. anonymous
    • one year ago
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    sine is a function, it is spelled "sine" if you want to write it in function notation, it needs to be \(\sin(\theta)\) or \(\sin(\xi)\) or \(\sin(\clubsuit)\) or something with a variable in the parentheses not "sin = " which makes no sense at all

  19. anonymous
    • one year ago
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    you can tell him/her satellite said they were abusing notation which will lead to massive confusion for you down the road

  20. anonymous
    • one year ago
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    Oh, ill be sure to tell her. What would i label the side lying on the x axix and the vertical side??

  21. anonymous
    • one year ago
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    x and y i guess

  22. anonymous
    • one year ago
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    |dw:1440642321127:dw|

  23. anonymous
    • one year ago
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    |dw:1440642400091:dw|

  24. anonymous
    • one year ago
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    |dw:1440642460037:dw|

  25. anonymous
    • one year ago
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    is that it?

  26. anonymous
    • one year ago
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    yeah looks good

  27. anonymous
    • one year ago
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    btw in the picture you have a \(\theta\) as the angle, so it should be \(\sin(\theta)\) etc also note that \(r=\sqrt{x^2+y^2}\) by pythagoras

  28. anonymous
    • one year ago
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    all right. Thanks. can you guide me with this?

  29. triciaal
    • one year ago
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    #6 use pythagoren theorem @satellite73 has it above

  30. anonymous
    • one year ago
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    yeah we just anticipated that question when i said \[r=\sqrt{x^2+y^2}\] so if, for example the point was \((2,5)\) you would have \[r=\sqrt{2^2+5^2}\]

  31. anonymous
    • one year ago
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    i notice on this page at least, the \(\theta\)'s have made an appearance !

  32. anonymous
    • one year ago
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    Lol. Would i leave it |dw:1440642870108:dw|

  33. anonymous
    • one year ago
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    keep the radical

  34. anonymous
    • one year ago
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    ok

  35. anonymous
    • one year ago
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    notice it says "no decimal"

  36. anonymous
    • one year ago
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    yeah. i just seen that

  37. anonymous
    • one year ago
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    So #7 would be |dw:1440643014145:dw|

  38. anonymous
    • one year ago
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    nvm

  39. anonymous
    • one year ago
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    lol

  40. anonymous
    • one year ago
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    |dw:1440643104712:dw|

  41. anonymous
    • one year ago
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    it is really \[\sin(\theta)=\frac{5}{\sqrt{29}}\] but that is not what they want as a "final answer"q

  42. anonymous
    • one year ago
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    yeah cause i cant leave the 29 like that.

  43. anonymous
    • one year ago
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    I would have to multiply ???

  44. anonymous
    • one year ago
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    aka "rationalize the denominator "

  45. anonymous
    • one year ago
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    \[\frac{5}{\sqrt{29}}\times \frac{\sqrt{29}}{\sqrt{29}}=\frac{5\sqrt{29}}{29}\] with a tiny bit of practice you do this in your head

  46. anonymous
    • one year ago
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    |dw:1440643207188:dw|

  47. anonymous
    • one year ago
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    there should be a 29 in the denominator

  48. anonymous
    • one year ago
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    So thats r??

  49. anonymous
    • one year ago
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    no \(r=\sqrt{29}\)

  50. anonymous
    • one year ago
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    \[\sin(\theta)=\frac{5\sqrt{29}}{29}\]

  51. anonymous
    • one year ago
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    hmmm so cos (theta)= 2/29?

  52. anonymous
    • one year ago
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    no \[\cos(\theta)=\frac{2}{\sqrt{29}}=\frac{2\sqrt{29}}{29}\]

  53. anonymous
    • one year ago
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    Tan = 5/2??

  54. anonymous
    • one year ago
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    yes

  55. anonymous
    • one year ago
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    nice. im doing #8 now

  56. anonymous
    • one year ago
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    ok find r as before, then use the same ratios as before

  57. anonymous
    • one year ago
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    ok

  58. anonymous
    • one year ago
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    the negatives are messing me up when im using the ratios

  59. anonymous
    • one year ago
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    Am i suppose to us positives??

  60. anonymous
    • one year ago
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    no

  61. triciaal
    • one year ago
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    plot the points and pay attention to the signs

  62. anonymous
    • one year ago
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    leave them as negative

  63. anonymous
    • one year ago
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    the tangent will be positive since you have two negatives, but the sine and cosine are negative for sure

  64. triciaal
    • one year ago
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    |dw:1440644210522:dw|

  65. anonymous
    • one year ago
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    So is this correct????

  66. triciaal
    • one year ago
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    |dw:1440644481593:dw|

  67. triciaal
    • one year ago
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    correct

  68. anonymous
    • one year ago
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    I got it correct??

  69. triciaal
    • one year ago
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    now you have the CAST rule

  70. triciaal
    • one year ago
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    |dw:1440644731475:dw|

  71. triciaal
    • one year ago
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    another mnemonic to remember what ratios are positive in which quadrant

  72. anonymous
    • one year ago
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    ok thanks

  73. triciaal
    • one year ago
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    A for all positive in quadrant 1

  74. triciaal
    • one year ago
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    you are welcome

  75. anonymous
    • one year ago
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    @satellite73

  76. anonymous
    • one year ago
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    @triciaal What about the S T C

  77. anonymous
    • one year ago
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  78. anonymous
    • one year ago
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    @satellite73

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is replying to Can someone tell me what button the professor is hitting...

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