Digital ACT Math Practice Question #2120
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Quadrant IV: Cosine is positive, Sine and Tangent are negative. 3-4-5 triangle $\implies \sin = -4/5$.
Since $\tan \theta 0$, $\theta$ must terminate in Quadrant IV. In Quadrant IV, $\sin \theta < 0$. With $\tan \theta = \frac{y}{x} = \frac{-4}{3}$, $r = \sqrt{3^2 + (-4)^2} = 5$. Therefore, $\sin \theta = \frac{y}{r} = -\frac{4}{5}$....
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