Question #a3f3d

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Answer 1 · 2017-08-18T14:34:13

$sin (θ) < 0$ $sin(theta)<0$

$tan (θ) =$ $tan(theta)=$ 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= $\sqrt{{(3)}^{2} + {(- 2)}^{2}} = \sqrt{5}$ $sqrt((3)^2+(-2)^2)=sqrt5$

So, $sin (θ) = - \frac{2}{\sqrt{5}}$ $sin(theta)=-2/sqrt5$

$sin (θ) < 0$ $sin(theta)<0$