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#