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anonymous

  • one year ago

Which of the following is true? Tangent is positive in Quadrant I. Sine is negative in Quadrant II. Cosine is positive in Quadrant III. Sine is positive in Quadrant IV.

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  1. anonymous
    • one year ago
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    @LexiLuvv2431 The true statement in that a tangent is positive in quadrant I.

  2. anonymous
    • one year ago
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    tangent is positive in first quadrant

  3. Nnesha
    • one year ago
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    |dw:1439494300583:dw| remember the \(\huge\color{green}{\rm CAST }\) rule

  4. anonymous
    • one year ago
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    Thank you guys so much I have a few more can you guys help?

  5. Nnesha
    • one year ago
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    |dw:1439494347347:dw|

  6. Nnesha
    • one year ago
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    |dw:1439494382810:dw|

  7. anonymous
    • one year ago
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    @LexiLuvv2431 u have ta give me da medal because I answerd it 1 lol

  8. anonymous
    • one year ago
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    Find the sine, cosine, and tangent of 120 degrees. and I did lol

  9. Nnesha
    • one year ago
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    use unit circle (x,y) where x -coordinate = cos y-coordinate = sin \[\rm tan \theta = \frac{ \sin \theta }{ \cos \theta }\]

  10. Nnesha
    • one year ago
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    |dw:1439494649288:dw|

  11. Nnesha
    • one year ago
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    so cos= ? sin=? tan = ?

  12. anonymous
    • one year ago
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    Sin is \[\frac{ \sqrt{3} }{ 2 }\]

  13. anonymous
    • one year ago
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    Idk the others though

  14. Nnesha
    • one year ago
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    hm yes that's right so if sin =sqrt{3}/2 cos = what look at the same parentheses where x-coordinate represents cos

  15. anonymous
    • one year ago
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    waittt okay so then would the cos equal 1/2 and the tan equal \[\sqrt{3}\] ?

  16. Nnesha
    • one year ago
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    cos is not just 1/2 look at first drawing i posted above cos is negative in which quadrant ?

  17. anonymous
    • one year ago
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    Quadrant S

  18. Nnesha
    • one year ago
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    |dw:1439495263173:dw| quadrant numbers i guess you don't understand why i wrote CAST C= cos S=sin T=tan A= all functions

  19. Nnesha
    • one year ago
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    |dw:1439495345327:dw| cos is negative in 2nd and 3rd quadrant

  20. Nnesha
    • one year ago
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    now look at the unit circle 120 degree is in 2nd quadrant where cos value should be negative |dw:1439495460867:dw|

  21. anonymous
    • one year ago
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    Noo sorry this is soo confusing for me.

  22. Nnesha
    • one year ago
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    just focus on it it's not hard you don't have to solve for anything just read the unit circle

  23. anonymous
    • one year ago
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    Ohhhh I seee. I get it nowww

  24. anonymous
    • one year ago
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    One more similar to this, The point (−3, 1) is on the terminal side of angle θ, in standard position. What are the values of sine, cosine, and tangent of θ? Make sure to show all work.

  25. Nnesha
    • one year ago
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    |dw:1439495526308:dw| it's like simple (x,y) graph

  26. Nnesha
    • one year ago
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    (x,y) x= cos y=sin so what is sin and cos in that equation ?

  27. anonymous
    • one year ago
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    -3 and 1??

  28. Nnesha
    • one year ago
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    yew which one is cos

  29. anonymous
    • one year ago
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    -3

  30. Nnesha
    • one year ago
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    yes right so you have draw a right triangle cos =-3 sin =1 sow at which quadrant you should draw a right triangle ?

  31. Nnesha
    • one year ago
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    cos = negative (-3) and sin =positive (1) hint!

  32. anonymous
    • one year ago
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    Quadrant 2

  33. Nnesha
    • one year ago
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    yep right !!

  34. Nnesha
    • one year ago
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    |dw:1439495963007:dw| apply Pythagorean theorem to find 3rd side of right triangle \[\huge\rm a^2+b^2=c^2\]

  35. Nnesha
    • one year ago
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    a=1 b=-3 solve for c

  36. anonymous
    • one year ago
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    100?

  37. Nnesha
    • one year ago
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    hmm 100 no how did you get that ?

  38. Nnesha
    • one year ago
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    replace a with 1 and b with -3 then take square of both numbers and then add them after that take square root

  39. anonymous
    • one year ago
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    ohhh I forgot the 3 was negative. So -64

  40. Nnesha
    • one year ago
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    show your work how did you get that ?

  41. anonymous
    • one year ago
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    I did 1*1 and got one and the -3 * -3 and got -9 and I added that and got -8 and then multiplied -8 * -8 and got -64

  42. Nnesha
    • one year ago
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    okay \[(1)+(-3)^2=c^2\] when you take even power of negative you will get positive answer

  43. Nnesha
    • one year ago
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    and i just noticed you said 100 earlier

  44. Nnesha
    • one year ago
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    which is correct but i thought you r saying c =100 c^2=100 take square root both sides you need c not c^2

  45. Nnesha
    • one year ago
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    wait no nvm

  46. anonymous
    • one year ago
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    Okay now I'm confused

  47. Nnesha
    • one year ago
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    \[\huge\rm (1)^2 + (-3)^2=c^2\]\[\rm 1+9=c^2\] (-3 times -3= 9) so now add 1+9\[\huge\rm 10=c^2\] now take square root both sides

  48. anonymous
    • one year ago
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    10 squared is 100

  49. Nnesha
    • one year ago
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    square root not square

  50. Nnesha
    • one year ago
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    to cancel out square you should take square root both sides to cancel out square root you should take square both sides so there is c^2 to cancel out square take square root

  51. anonymous
    • one year ago
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    Okay I have no idea how to do this part

  52. Nnesha
    • one year ago
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    \[\huge\rm \sqrt{c^2}= \sqrt{10}\]take square root both sides sqrt{c^2} = c so \[c =\sqrt{10}\]

  53. Nnesha
    • one year ago
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    |dw:1439497283978:dw|

  54. anonymous
    • one year ago
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    Okayy I see

  55. Nnesha
    • one year ago
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    now remember definition of sin,cos and tan you need this \[\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }\]

  56. Nnesha
    • one year ago
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    |dw:1439497367283:dw| opp= opposite side of theta adj= adjacent of theta hyp=hypotenuse of theta

  57. anonymous
    • one year ago
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    |dw:1439508247714:dw|

  58. anonymous
    • one year ago
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    So is the answer finished?

  59. Nnesha
    • one year ago
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    yes right just write value of sin cosine and tan

  60. anonymous
    • one year ago
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    How do I do that?

  61. Nnesha
    • one year ago
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    \[\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }\] just replace opposite , adjacent hyp words with their values

  62. Nnesha
    • one year ago
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    |dw:1439498158642:dw|

  63. anonymous
    • one year ago
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    Thank youu so much

  64. Nnesha
    • one year ago
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    like sin = opposite over hyp so opposite side of theta is 1 and hyp = sqrt{10} write it as \[\sin \theta =\frac{ 1 }{ \sqrt{10} }\] you can't leave the radical sign at the denominator so multiply top and bottom by square root 10

  65. Nnesha
    • one year ago
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    my pleasure:)

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