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rebeccaskell94

  • 3 years ago

Given the triangle below, what is cot∡B? http://learn.flvs.net/webdav/assessment_images/educator_geometry_v14/pool_Geom_3641_0403_03/image0084e2ec0b0.gif √33/ 4 7•√33/ 33 √33 /7 4•√33/33 I think the Answer is C?

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  1. goformit100
    • 3 years ago
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    Use Pythagoras theorem

  2. mathslover
    • 3 years ago
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    I think that is a) \(\sqrt{33}/4\)

  3. rebeccaskell94
    • 3 years ago
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    urg I couldn't figure out if it was a or c because I wasn't sure if 4 or 7 should be the denominator

  4. mathslover
    • 3 years ago
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    cot B = \(\frac{1}{tanB}\) right ?

  5. ash2326
    • 3 years ago
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    @rebeccaskell94 First step is to find AB we know that using Pythagoras \[BC^2=AB^2+AC^2\] Find AB from here \[\cot =\frac{adjacent}{perpendicular}=\frac1 {\tan}\]

  6. mathslover
    • 3 years ago
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    tan B = opposite/adjacent opposite = 4 adjacent = ??? We will find adj. by pythagoras theorem

  7. mathslover
    • 3 years ago
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    \[base^2+perpendicular^2=hypotenuse^2\] Adjacent = \(\sqrt{33}\) hence we have tan B = \(\frac{\sqrt{33}}{4}\)

  8. mathslover
    • 3 years ago
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    did it help @rebeccaskell94 ?

  9. rebeccaskell94
    • 3 years ago
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    Yes! Very much so. I'm writing it down now so I remember the basic idea :) Thank you so much!

  10. mathslover
    • 3 years ago
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    Nice to hear that @rebeccaskell94 :)

  11. phi
    • 3 years ago
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    See http://www.khanacademy.org/math/trigonometry/v/basic-trigonometry for a video on sin,cos, tan defs using SOHCAHTOA a the co functions are harder to remember!

  12. mathslover
    • 3 years ago
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    SOH CAH TOA much better sine = opposite / hypotenuse cosine = adj. / hypotenuse tan = opposite / adj. right ?

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