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Trigonometry help? Has to do with unit circles.
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\[\theta=\pi/6\]
sorry, \[\theta=\pi/3\]
how did you get that answer?
tan=sin/cos, at \[\pi/3\] \[(\frac{ 1 }{ 2 },\frac{ \sqrt{3} }{ 2 })\] (cos, sin)
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\[\frac{ \frac{ \sqrt{3} }{ 2 } }{ \frac{ 1 }{ 2 } }=\sqrt{3}\]
alternately, \[\tan^{-1} (\sqrt{3})\]
ahh so it hits the point on a 30-60-90 triangle
right, drop a vertical perpendicular line from the end point to the x axis
|dw:1361902605164:dw| would look something like this?
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