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√3 csc θ - 2 = 0
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\[cosec \theta =2/\sqrt{3}\] thus \[\theta =60\]
ok \[\csc (\theta)=\frac{ 2 }{ \sqrt{3} }\] \[\csc(\theta)=\frac{ a }{ b }\] \[\sin(\theta)=\frac{ b }{ a }\]
so here we have \[\sin (\theta)=\frac{ \sqrt{3} }{ 2 }\]
to find theta we can either look at a unit circle or do a sin^-1
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remember \[\sin (\theta) = \frac{ opposite }{ hypotenuse }\]
and in this case we have a positive value
so our reference angles are going to be in quadrants 1 and 2 where sin (the opposite side) is positive like so |dw:1408171441298:dw|
|dw:1408171477971:dw|
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