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anonymous

  • one year ago

Assuming cos t = .45 and cos w = .89, both t and w are positive, and both t and w determine a terminal point in quadrant 1, then which of the following statements best describes the relationship between t and w? A. t > w B. w > t C. It is not possible to tell from the given information.

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  1. freckles
    • one year ago
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    Have you looked at the first quadrant and notice what happens as the cos(t) values increase? like what happens to t?

  2. freckles
    • one year ago
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    you are given t and w are positive

  3. freckles
    • one year ago
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    so yes

  4. freckles
    • one year ago
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    but that doesn't answer your question

  5. freckles
    • one year ago
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    https://www.mathsisfun.com/geometry/images/circle-unit-304560.gif Look at this picture what is cos(60) and cos(30) equal to?

  6. anonymous
    • one year ago
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    cos(60) = 1/2 and cos(30) = \[\sqrt{3}/2\]

  7. freckles
    • one year ago
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    \[60^o >30^o \\ \cos(60^o)=\frac{1}{2} ? \frac{\sqrt{3}}{2}=\cos(30^o)\] which of those values are greater ?

  8. freckles
    • one year ago
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    is 1 bigger than sqrt(3) or is sqrt(3) bigger than 1?

  9. anonymous
    • one year ago
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    sqrt(3) is bigger than 1

  10. freckles
    • one year ago
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    \[60^o>30^o \\ \cos(60^o)<\cos(30^o) \\ \text{ and you have } \cos(t)<\cos(w)\] and since we know t and w are in the first quadrant just like 60deg and 30deg was then we know what about t and w?

  11. anonymous
    • one year ago
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    We know that t < w?

  12. freckles
    • one year ago
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    why? didn't we have that cos(60)<cos(30) but 60>30 ?

  13. anonymous
    • one year ago
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    Was it because of the cosine?

  14. freckles
    • one year ago
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    what does that mean?

  15. anonymous
    • one year ago
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    The cosine of 60 is giving us a value that is less than our cosine of 30 I think

  16. freckles
    • one year ago
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    yes 1/2 is less than sqrt(3)/2

  17. anonymous
    • one year ago
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    So the answer would be w > t (B)?

  18. freckles
    • one year ago
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    ok see if you follow this: \[60^o>30^o \\ \text{ so } \cos(60^o)=\frac{1}{2}<\frac{\sqrt{3}}{2}=\cos(30^o) \\ \text{ we have} \cos(60^o)<\cos(30^o) \text{ but } 60^o>30^o \\ \text{ You are given } \cos(t)<\cos(w) \text{ so you can draw what conclusion about } t?w\] hint replace 60 with t and 30 with w

  19. freckles
    • one year ago
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    you can also notice in the first quadrant as the angle increases the cos value of those angles decrease

  20. anonymous
    • one year ago
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    So if the equations decreased that would make t > w?

  21. anonymous
    • one year ago
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    Great! Thank you!!!

  22. freckles
    • one year ago
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    yes

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