1. **Problem Statement:** Given the terminal side of angle $\theta$ lies on the line $-4x + 3y = 0$ in Quadrant IV, find the values of $\sin \theta$, $\cos \theta$, $\tan \theta$, $\sec \theta$, and $\csc \theta$. 2. **Rewrite the line equation:** From $-4x + 3y = 0$, solve for $y$: $$3y = 4x \implies y = \frac{4}{3}x$$ 3. **Determine the coordinates on the terminal side:** Since the terminal side lies on this line, any point $(x,y)$ satisfies $y = \frac{4}{3}x$. 4. **Identify the quadrant and signs:** Quadrant IV means $x > 0$ and $y < 0$. 5. **Adjust signs to fit Quadrant IV:** Since $y = \frac{4}{3}x$ is positive for $x > 0$, to be in Quadrant IV where $y < 0$, we take $y = -\frac{4}{3}x$. 6. **Find the radius $r$ (distance from origin):** $$r = \sqrt{x^2 + y^2} = \sqrt{x^2 + \left(-\frac{4}{3}x\right)^2} = \sqrt{x^2 + \frac{16}{9}x^2} = \sqrt{\frac{25}{9}x^2} = \frac{5}{3}|x|$$ Since $x > 0$, $|x| = x$, so: $$r = \frac{5}{3}x$$ 7. **Calculate trigonometric ratios:** - $$\sin \theta = \frac{y}{r} = \frac{-\frac{4}{3}x}{\frac{5}{3}x} = -\frac{4}{5}$$ - $$\cos \theta = \frac{x}{r} = \frac{x}{\frac{5}{3}x} = \frac{3}{5}$$ - $$\tan \theta = \frac{y}{x} = -\frac{4}{3}$$ - $$\sec \theta = \frac{1}{\cos \theta} = \frac{1}{\frac{3}{5}} = \frac{5}{3} \approx 1.6667$$ - $$\csc \theta = \frac{1}{\sin \theta} = \frac{1}{-\frac{4}{5}} = -\frac{5}{4} = -1.25$$ 8. **Convert to decimals as requested:** - $\sin \theta = -0.8$ - $\cos \theta = 0.6$ - $\tan \theta = -1.3333$ - $\sec \theta = 1.6667$ - $\csc \theta = -1.25$ **Final answers:** $$\sin \theta = -0.8$$ $$\cos \theta = 0.6$$ $$\tan \theta = -1.3333$$ $$\sec \theta = 1.6667$$ $$\csc \theta = -1.25$$