How to solve: tan(theta) = 2/(-2 sqroot3)
\[\tan(\theta)=\frac{ 2 }{ -2\sqrt{3} }\]
\[\theta = \tan ^{-1}(\frac{ 2 }{ -2\sqrt{3} })\]
@Nnesha
@jim_thompson5910
can you cancel out 2's ??
multiply it by 2/2?
ooh I see what you're saying, sorry.
err can I?
then it would be \[\frac{ 1 }{ -\sqrt{3} }\]
\[\huge\rm tan(\theta) =\color{red}{-\frac{ 2 }{ 2\sqrt{3}} }\] first of all solve red part :-)
\(\color{blue}{\text{Originally Posted by}}\) @Babynini then it would be \[\frac{ 1 }{ -\sqrt{3} }\] \(\color{blue}{\text{End of Quote}}\) yes but you ca't have radical at the denominator so multiply top and bottom by sqrt{3}
\[\frac{ \sqrt{3} }{ 3 }\]
with a negative in front :P
\[\theta= \tan^{-1} (-\frac{\sqrt{3}}{3} )\] from there do you wanna use a calculator or draw triangle ?
Triangle! Because what I need is it in radical form
(for example 5pi/6)
|dw:1432009784662:dw| to make a triangle we know that it's not going to be in 1st and 3rd quadrant bec tan is positive in 1st and 3rd q
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