1. **State the problem:** Find the angle between the line AG and the plane ABCD in the cuboid with edges 15.3 cm, 9.5 cm, and 8.3 cm. 2. **Identify the dimensions and points:** - Let the cuboid have base ABCD with AB = 15.3 cm, BC = 9.5 cm, and height AE = 8.3 cm. - Point G is the vertex diagonally opposite A on the top face, so AG is the space diagonal from A to G. 3. **Find vector AG:** - Vector AG = AB + BC + AE = (15.3, 9.5, 8.3) in coordinate form. 4. **Find the normal vector to plane ABCD:** - Plane ABCD lies in the xy-plane, so its normal vector is along the z-axis: $\vec{n} = (0,0,1)$. 5. **Formula for angle between a line and a plane:** - The angle $\theta$ between line and plane satisfies $\sin \theta = \frac{|\vec{d} \cdot \vec{n}|}{|\vec{d}| |\vec{n}|}$, where $\vec{d}$ is the direction vector of the line. - Alternatively, the angle between the line and plane is $90^\circ - \phi$, where $\phi$ is the angle between the line and the normal vector. 6. **Calculate magnitude of AG:** $$|\vec{AG}| = \sqrt{15.3^2 + 9.5^2 + 8.3^2} = \sqrt{234.09 + 90.25 + 68.89} = \sqrt{393.23} \approx 19.83$$ 7. **Calculate dot product of AG and normal vector:** $$\vec{AG} \cdot \vec{n} = 0 \times 15.3 + 0 \times 9.5 + 1 \times 8.3 = 8.3$$ 8. **Calculate sine of angle between line and plane:** $$\sin \theta = \frac{|8.3|}{19.83 \times 1} = \frac{8.3}{19.83} \approx 0.4185$$ 9. **Calculate angle $\theta$:** $$\theta = \arcsin(0.4185) \approx 24.8^\circ$$ 10. **Final answer:** The angle between line AG and plane ABCD is approximately **25°** to the nearest degree.