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Amenah8

  • one year ago

FIND FAETA: cos faeta = (-15/17), sin faeta > 0

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  1. Michele_Laino
    • one year ago
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    do you mean cos(\theta) and sin(\theta), right?

  2. Amenah8
    • one year ago
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    Right, sorry!

  3. Michele_Laino
    • one year ago
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    the requested angle \theta lies in the second quadrant, since we have this drawing: |dw:1439224267236:dw|

  4. Michele_Laino
    • one year ago
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    |dw:1439224370587:dw|

  5. Amenah8
    • one year ago
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    how did you know it was the 2nd quadrant?

  6. Michele_Laino
    • one year ago
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    sionce if the point P, namely the second ends of your arc, lies in the second quadrant, then its x-coordinate, which is equal to cos(\theta) is negative, whereas its y-coordinate, which is sin(\theta) is positive, as requested from your problem

  7. Michele_Laino
    • one year ago
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    since*

  8. Amenah8
    • one year ago
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    oh, i get it!

  9. Michele_Laino
    • one year ago
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    so you have to find an angle whose measure is greater than 90 degrees and less than 180 degrees, whose cosine function is equal to -15/17

  10. Amenah8
    • one year ago
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    and i would use the graphing circle

  11. Michele_Laino
    • one year ago
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    for computation, you can use, for example, windows calculator

  12. Amenah8
    • one year ago
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    but if cos is just x, x=-15/7. so then how do we know y (sin) is positive? couldn't it have been in the 3rd quadrant?

  13. Michele_Laino
    • one year ago
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    sin(\theta) has to be positive, since it is a data of your problem

  14. Amenah8
    • one year ago
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    what do you mean by a data of my problem?

  15. Michele_Laino
    • one year ago
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    |dw:1439224865101:dw|

  16. Michele_Laino
    • one year ago
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    as we can see from my last drawing, only points which belongs to the first quadrant and second quadrant, have y-coordinate >0

  17. Michele_Laino
    • one year ago
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    belong*

  18. Amenah8
    • one year ago
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    oh, because in third and 4th quadrants, y is negative!

  19. Michele_Laino
    • one year ago
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    yes!

  20. Amenah8
    • one year ago
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    so how would i solve for theta? on the unit circle, the only x "coordinates" are the negative square root of 3 divided by 2, -1/2, and the negative square root of 2 divided by 2. None of those exactly equal -15/17.

  21. Amenah8
    • one year ago
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    but -\[-\sqrt{3}/2\] is close

  22. Michele_Laino
    • one year ago
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    you have to find \theta, right?

  23. Michele_Laino
    • one year ago
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    if cos(\theta)= -15/17, then using windows calculator I got: \theta= 151.93 degrees, namely \theta= 152 degrees

  24. Amenah8
    • one year ago
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    how did you get 151.93?

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