1. **Problem statement:** Given point $P(-3, y)$ lies on a circle centered at the origin $O$ with radius $OP = 5$ units, and angle $\alpha = \angle XÔP$. 2. **Find $y$:** Use the distance formula for point $P$ from origin $O$: $$OP = \sqrt{(-3)^2 + y^2} = 5$$ Square both sides: $$(-3)^2 + y^2 = 5^2$$ $$9 + y^2 = 25$$ Subtract 9: $$y^2 = 16$$ Take square root: $$y = \pm 4$$ Since $P$ is above or below x-axis, $y$ can be $4$ or $-4$. 3. **Find $\sin \alpha$:** $\sin \alpha = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{y}{5}$ So, $$\sin \alpha = \frac{4}{5} \text{ or } \frac{-4}{5}$$ 4. **Find $\cos(180^\circ - \alpha)$:** Use the identity: $$\cos(180^\circ - \alpha) = -\cos \alpha$$ Calculate $\cos \alpha = \frac{x}{5} = \frac{-3}{5}$ So, $$\cos(180^\circ - \alpha) = -\left(\frac{-3}{5}\right) = \frac{3}{5}$$ 5. **Find $\alpha$ (using calculator):** $$\cos \alpha = \frac{-3}{5} = -0.6$$ So, $$\alpha = \cos^{-1}(-0.6) = 126.87^\circ$$ 6. **Find area of $\triangle POX$ where $X(2,0)$:** Use formula for area of triangle with vertices $P(-3,4)$, $O(0,0)$, $X(2,0)$: $$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$ Substitute: $$= \frac{1}{2} |-3(0 - 0) + 0(0 - 4) + 2(4 - 0)|$$ $$= \frac{1}{2} |0 + 0 + 8| = 4$$ **Final answers:** - $y = \pm 4$ - $\sin \alpha = \pm \frac{4}{5}$ - $\cos(180^\circ - \alpha) = \frac{3}{5}$ - $\alpha = 126.87^\circ$ - Area of $\triangle POX = 4$ units squared