Digital ACT Math Practice Question #1209
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Remember 'All Students Take Calculus' (ASTC). Tangent is positive in QIII, where both Sine and Cosine are negative. Cosine MUST be negative, eliminating B and D instantly. 5-12-13 triangle gives $-12/13$.
Since $\sin(\theta) 0$, the angle $\theta$ must terminate in Quadrant III. In Quadrant III, cosine is negative. Using the Pythagorean identity: $\cos^2(\theta) = 1 - \sin^2(\theta) = 1 - (-5/13)^2 = 1 - 25/169 = 144/169$. Since $\theta$ is ...
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