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The line with equation a + 2b = 0 coincides with the terminal side of an angle θ in standard position and cos θ<0.

What is the value of sinθ?

 

−2√5/5

√5

−1/2

√5/5

Guest Dec 8, 2017
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1a + 2b  = 0

 

2b  =  -1a

 

b  =  (-1/2)a

 

The slope of this line is    (-1/2)....this line lies in the II and IV  quadrants...and since the cos < 0, then this must be a II quadrant angle so the sine must > 0

 

So.....find r  = √ [ x^2 + y^2 ]  = √  ] (-2)^2 + (1)^2 ]  =  √5

 

So....the sinθ  =  1/ √5  =  √5 / 5

 

 

cool cool cool

CPhill  Dec 8, 2017

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