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highschoolmom2010

  • 2 years ago

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

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  1. katherine.ok
    • 2 years ago
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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?

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

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

  4. katherine.ok
    • 2 years ago
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    So the normal numerator of sin would be denomenator of csc , and etc..

  5. highschoolmom2010
    • 2 years ago
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    so its like flipped

  6. tkhunny
    • 2 years ago
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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. \(\sin(B) = \dfrac{m(Opposite\;Side)}{m(Hypotenuse)} = \dfrac{1}{\csc(B)}\) \(\cos(B) = \dfrac{m(Adjacent\;Side)}{m(Hypotenuse)} = \dfrac{1}{\sec(B)}\) \(\tan(B) = \dfrac{m(Opposite\;Side)}{m(Adjacent\;Side)} = \dfrac{1}{\cot(B)}\)

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

  8. highschoolmom2010
    • 2 years ago
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    cos A= 9/15

  9. tkhunny
    • 2 years ago
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    Yes!!!

  10. highschoolmom2010
    • 2 years ago
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    so would the sec A= 15/9?

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

  12. highschoolmom2010
    • 2 years ago
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    sin (b)=9/15

  13. highschoolmom2010
    • 2 years ago
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    csc=15/9

  14. agent0smith
    • 2 years ago
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    Correct. Now try cotB. cotB = 1/tanB.

  15. highschoolmom2010
    • 2 years ago
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    |dw:1376265085663:dw| \[\tan (B)=\frac{ 9 }{ 12 }\]

  16. highschoolmom2010
    • 2 years ago
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    cot B= 12/9

  17. highschoolmom2010
    • 2 years ago
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    @agent0smith do i have to simplify ^^

  18. agent0smith
    • 2 years ago
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    Yes, I would.

  19. highschoolmom2010
    • 2 years ago
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    so the cot (b)= 4/3 csc (B)=5/3 sec (B)=5/3

  20. agent0smith
    • 2 years ago
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    Yep :)

  21. highschoolmom2010
    • 2 years ago
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    cool i get it now :D ty

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

  23. highschoolmom2010
    • 2 years ago
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    @agent0smith i have A written down lol

  24. highschoolmom2010
    • 2 years ago
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    ty @agent0smith @tkhunny & @katherine.ok for the help :D

  25. agent0smith
    • 2 years ago
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    Good job :)

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