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highschoolmom2010 Group Title

The sine, cosine, and tangent ratios each have a reciprocal ratio. The reciprocal ratios are cosecant (csc), secant (sec), and cotangent (cot). use triangle ABC to write the definitions sec A csc B cot B

  • 11 months ago
  • 11 months ago

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  1. katherine.ok Group Title
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    Are you given or have you learned that csc= reciprocal of which , and sec of which , and cot is reciprocal of which?

    • 11 months ago
  2. highschoolmom2010 Group Title
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    |dw:1376262907756:dw| not covered in my book so i kinda know a little

    • 11 months ago
  3. katherine.ok Group Title
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    csc= sin inverse. cot= tan inverse, sec= cos inverse. (Sorry for the informal uses of words)

    • 11 months ago
  4. katherine.ok Group Title
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    So the normal numerator of sin would be denomenator of csc , and etc..

    • 11 months ago
  5. highschoolmom2010 Group Title
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    so its like flipped

    • 11 months ago
  6. tkhunny Group Title
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    For the sake of clarity, let's not use "inverse" in this manner. This is a reciprocal. I can see why you used "inverse", but a few lessons down the road, you WILL regret it if you call it that. Pick an Angle!! Let's call it "B". It your drawing, it's one of the acute angles. After that, the definitions require memorization. You will be required to know which side is adjacent to the given acute angle and which side is opposite the given acute angle. \(\sin(B) = \dfrac{m(Opposite\;Side)}{m(Hypotenuse)} = \dfrac{1}{\csc(B)}\) \(\cos(B) = \dfrac{m(Adjacent\;Side)}{m(Hypotenuse)} = \dfrac{1}{\sec(B)}\) \(\tan(B) = \dfrac{m(Opposite\;Side)}{m(Adjacent\;Side)} = \dfrac{1}{\cot(B)}\)

    • 11 months ago
  7. agent0smith Group Title
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    SecA = 1/cosA Find cosA first, this will let you find sec A. CosA = adjacent/hypotenuse. |dw:1376263854146:dw|

    • 11 months ago
  8. highschoolmom2010 Group Title
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    cos A= 9/15

    • 11 months ago
  9. tkhunny Group Title
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    Yes!!!

    • 11 months ago
  10. highschoolmom2010 Group Title
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    so would the sec A= 15/9?

    • 11 months ago
  11. agent0smith Group Title
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    Correct! Now try cscB = 1/sinB Sin = opp/hyp. |dw:1376264245185:dw|

    • 11 months ago
  12. highschoolmom2010 Group Title
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    sin (b)=9/15

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  13. highschoolmom2010 Group Title
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    csc=15/9

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  14. agent0smith Group Title
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    Correct. Now try cotB. cotB = 1/tanB.

    • 11 months ago
  15. highschoolmom2010 Group Title
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    |dw:1376265085663:dw| \[\tan (B)=\frac{ 9 }{ 12 }\]

    • 11 months ago
  16. highschoolmom2010 Group Title
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    cot B= 12/9

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  17. highschoolmom2010 Group Title
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    @agent0smith do i have to simplify ^^

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  18. agent0smith Group Title
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    Yes, I would.

    • 11 months ago
  19. highschoolmom2010 Group Title
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    so the cot (b)= 4/3 csc (B)=5/3 sec (B)=5/3

    • 11 months ago
  20. agent0smith Group Title
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    Yep :)

    • 11 months ago
  21. highschoolmom2010 Group Title
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    cool i get it now :D ty

    • 11 months ago
  22. agent0smith Group Title
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    Oh except wasn't one of those sec A? secA was 5/3

    • 11 months ago
  23. highschoolmom2010 Group Title
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    @agent0smith i have A written down lol

    • 11 months ago
  24. highschoolmom2010 Group Title
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    ty @agent0smith @tkhunny & @katherine.ok for the help :D

    • 11 months ago
  25. agent0smith Group Title
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    Good job :)

    • 11 months ago
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