1. **Problem Statement:** Find the angle between the diagonal $AG$ and the plane $EFGH$. 2. **Understanding the prism and coordinates:** Assign coordinates to vertices for clarity: - $A = (0,0,0)$ - $B = (4,0,0)$ (since $AB=4$) - $C = (4,4,0)$ (since $BC=4$) - $D = (0,4,0)$ - $E = (0,0,8)$ (vertical edge $FG=8$) - $F = (4,0,8)$ - $G = (4,4,8)$ - $H = (0,4,8)$ 3. **Vectors involved:** - Diagonal $AG = G - A = (4,4,8)$ - Plane $EFGH$ is the top face with points $E, F, G, H$ all at $z=8$. 4. **Find the normal vector to plane $EFGH$:** Since $EFGH$ is horizontal at $z=8$, its normal vector is vertical: $$\vec{n}_{EFGH} = (0,0,1)$$ 5. **Angle between line and plane:** The angle $\theta$ between a line and a plane is related to the angle $\phi$ between the line's direction vector and the plane's normal vector by: $$\theta = 90^\circ - \phi$$ where $$\cos \phi = \frac{|\vec{AG} \cdot \vec{n}_{EFGH}|}{|\vec{AG}| |\vec{n}_{EFGH}|}$$ 6. **Calculate magnitudes and dot product:** $$|\vec{AG}| = \sqrt{4^2 + 4^2 + 8^2} = \sqrt{16 + 16 + 64} = \sqrt{96} = 4\sqrt{6}$$ $$|\vec{n}_{EFGH}| = 1$$ $$\vec{AG} \cdot \vec{n}_{EFGH} = 4 \times 0 + 4 \times 0 + 8 \times 1 = 8$$ 7. **Calculate $\cos \phi$ and then $\theta$:** $$\cos \phi = \frac{8}{4\sqrt{6}} = \frac{8}{4\sqrt{6}} = \frac{2}{\sqrt{6}}$$ $$\phi = \cos^{-1}\left(\frac{2}{\sqrt{6}}\right)$$ 8. **Calculate $\theta$:** $$\theta = 90^\circ - \phi = 90^\circ - \cos^{-1}\left(\frac{2}{\sqrt{6}}\right)$$ 9. **Numerical approximation:** $$\frac{2}{\sqrt{6}} \approx 0.8165$$ $$\phi \approx \cos^{-1}(0.8165) \approx 35^\circ$$ $$\theta \approx 90^\circ - 35^\circ = 55^\circ$$ **Final answer:** The angle between diagonal $AG$ and plane $EFGH$ is approximately $55^\circ$.