NicholasWong752
Evaluate. ArcTan( sqrt3/ 3)
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NicholasWong752
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\[\tan^{-1} \frac{ \sqrt{3} }{ 3 }\]
geerky42
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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 \( \dfrac{\sqrt3}{3} \)?
NicholasWong752
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It's 30. But I don't know how I got that.
NicholasWong752
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|dw:1357522052726:dw|
geerky42
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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|
NicholasWong752
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I thought Tangent was opposite over adjacent.
NicholasWong752
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OH THEY RATIONALIZED IT
geerky42
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It is.
Welcome.
NicholasWong752
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God damn it, tricky bastards. >:|
Thanks a lot. xD