Question #a3f3d

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Answer 1 · 2017-08-18T14:34:13

$sin(theta)<0$

$tan(theta)=$ opposite $/$ adjacent

So, let's think about triangle in x and y coordinate

opposite = -2
adjacent = 3

so that triangle will be in the fourth part in the x and y coordinate

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hypotenuse= $sqrt((3)^2+(-2)^2)=sqrt5$

So, $sin(theta)=-2/sqrt5$

$sin(theta)<0$