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NicholasWong752

  • one year ago

Evaluate. ArcTan( sqrt3/ 3)

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  1. NicholasWong752
    • one year ago
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    \[\tan^{-1} \frac{ \sqrt{3} }{ 3 }\]

  2. geerky42
    • one year ago
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    Let \(\theta\) be \( \tan^{-1} \dfrac{\sqrt{3}}{3} \). \(\tan \theta = \dfrac{\sqrt{3}}{3}\) At what angle is \(\tan \theta\) equal to \( \dfrac{\sqrt3}{3} \)?

  3. NicholasWong752
    • one year ago
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    It's 30. But I don't know how I got that.

  4. NicholasWong752
    • one year ago
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    |dw:1357522052726:dw|

  5. geerky42
    • one year ago
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    Well, \(\tan 30^{\text{o}} = \dfrac{\sqrt3}{3}\) so it's true that \(30^{\text{o}} = \tan^{-1}\dfrac{\sqrt3}{3}\).|dw:1357522138906:dw|

  6. NicholasWong752
    • one year ago
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    I thought Tangent was opposite over adjacent.

  7. NicholasWong752
    • one year ago
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    OH THEY RATIONALIZED IT

  8. geerky42
    • one year ago
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    It is. Welcome.

  9. NicholasWong752
    • one year ago
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    God damn it, tricky bastards. >:| Thanks a lot. xD

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