1. **Problem Statement:** Given the terminal side of angle $\theta$ lies on the line $-4x + 3y = 0$ in Quadrant IV, find $\sin \theta$, $\cos \theta$, $\tan \theta$, $\sec \theta$, and $\csc \theta$ in decimal form. 2. **Rewrite the line equation:** From $-4x + 3y = 0$, solve for $y$: $$3y = 4x \implies y = \frac{4}{3}x$$ 3. **Choose a point on the line in Quadrant IV:** In Quadrant IV, $x > 0$ and $y < 0$. Since $y = \frac{4}{3}x$, to have $y < 0$, $x$ must be positive and $y$ negative. So choose $x = 3$ (positive), then: $$y = \frac{4}{3} \times 3 = 4$$ But $y$ is positive here, so to get $y < 0$, choose $x = 3$ and $y = -4$ (negate $y$ to be in Quadrant IV). This point $(3, -4)$ lies on the line if we check: $$-4(3) + 3(-4) = -12 - 12 = -24 \neq 0$$ So we must keep the original line equation and find a point with $y = \frac{4}{3}x$ but $y < 0$. Since $y = \frac{4}{3}x$, if $x$ is negative, $y$ is negative, but that would be Quadrant III. To be in Quadrant IV, $x > 0$ and $y < 0$, but $y = \frac{4}{3}x$ is positive if $x > 0$. So the line $y = \frac{4}{3}x$ lies in Quadrants I and III only. However, the original line is $-4x + 3y = 0$, or $y = \frac{4}{3}x$. The terminal side lies on this line in Quadrant IV, so the point must satisfy the line and be in Quadrant IV. To satisfy both, consider the vector $(x,y)$ proportional to $(3,-4)$ because the line can also be written as $4x - 3y = 0$ or $y = \frac{4}{3}x$, but the terminal side can be in Quadrant IV if we take the vector $(3,-4)$ which satisfies: $$-4(3) + 3(-4) = -12 - 12 = -24 \neq 0$$ So check the original line carefully: Rewrite $-4x + 3y = 0$ as $3y = 4x$ or $y = \frac{4}{3}x$. If $x > 0$, then $y > 0$, so the line passes through Quadrants I and III only. Therefore, the terminal side cannot lie on this line in Quadrant IV unless the problem means the line $4x - 3y = 0$ (which would be $y = \frac{4}{3}x$) or the negative of the vector. Assuming the terminal side vector is $(3,-4)$ (which lies in Quadrant IV), check if it satisfies the line: $$-4(3) + 3(-4) = -12 - 12 = -24 \neq 0$$ No. Try $( -3, 4 )$: $$-4(-3) + 3(4) = 12 + 12 = 24 \neq 0$$ No. Try $(4,3)$: $$-4(4) + 3(3) = -16 + 9 = -7 \neq 0$$ No. Try $( -4, -3 )$: $$-4(-4) + 3(-3) = 16 - 9 = 7 \neq 0$$ No. Try $(3,4)$: $$-4(3) + 3(4) = -12 + 12 = 0$$ Yes, but this is Quadrant I. Try $( -3, -4 )$: $$-4(-3) + 3(-4) = 12 - 12 = 0$$ Yes, Quadrant III. So the line passes through Quadrants I and III only. Since the problem states Quadrant IV, the terminal side must be the vector opposite to $(3,4)$, i.e., $(3,-4)$, but this does not satisfy the line. Hence, the problem likely means the line $4x - 3y = 0$ or $-4x - 3y = 0$. Assuming the line is $4x - 3y = 0$, then $y = \frac{4}{3}x$. For Quadrant IV, $x > 0$, $y < 0$, so $y = -\frac{4}{3}x$. Therefore, the vector is proportional to $(3,-4)$. 4. **Calculate $r$ (radius):** $$r = \sqrt{x^2 + y^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$ 5. **Calculate trigonometric ratios:** - $\sin \theta = \frac{y}{r} = \frac{-4}{5} = -0.8$ - $\cos \theta = \frac{x}{r} = \frac{3}{5} = 0.6$ - $\tan \theta = \frac{y}{x} = \frac{-4}{3} \approx -1.3333$ - $\sec \theta = \frac{1}{\cos \theta} = \frac{1}{0.6} \approx 1.6667$ - $\csc \theta = \frac{1}{\sin \theta} = \frac{1}{-0.8} = -1.25$ **Final answers:** $$\sin \theta = -0.8$$ $$\cos \theta = 0.6$$ $$\tan \theta \approx -1.3333$$ $$\sec \theta \approx 1.6667$$ $$\csc \theta = -1.25$$ These values correspond to the terminal side lying on the line $4x - 3y = 0$ in Quadrant IV.