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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

  • one year ago
  • one year 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?

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

    • one year 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)

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

    • one year ago
  5. highschoolmom2010 Group Title
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    so its like flipped

    • one year 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. \(\sin(B) = \dfrac{m(Opposite\;Side)}{m(Hypotenuse)} = \dfrac{1}{\csc(B)}\) \(\cos(B) = \dfrac{m(Adjacent\;Side)}{m(Hypotenuse)} = \dfrac{1}{\sec(B)}\) \(\tan(B) = \dfrac{m(Opposite\;Side)}{m(Adjacent\;Side)} = \dfrac{1}{\cot(B)}\)

    • one year 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|

    • one year ago
  8. highschoolmom2010 Group Title
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    cos A= 9/15

    • one year ago
  9. tkhunny Group Title
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    Yes!!!

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

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

    • one year 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.

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

    • one year 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.

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

    • one year ago
  20. agent0smith Group Title
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    Yep :)

    • one year ago
  21. highschoolmom2010 Group Title
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    cool i get it now :D ty

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

    • one year ago
  23. highschoolmom2010 Group Title
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    @agent0smith i have A written down lol

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

    • one year ago
  25. agent0smith Group Title
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    Good job :)

    • one year ago
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