1. **State the problem:** We have a cuboid ABCDEFGH with length $17$ cm, width $5$ cm, and height $8$ cm. We need to find the angle between the line segment $AH$ and the plane $EFGH$. 2. **Identify the vectors:** The plane $EFGH$ lies on the base of the cuboid, so it is parallel to the $xy$-plane. The vector $\overrightarrow{AH}$ runs from point $A$ to point $H$. 3. **Coordinates of points:** - Let $A$ be at the origin $(0,0,0)$. - Since $AB$ is height $8$ cm, $B$ is at $(0,0,8)$. - $H$ is at $(17,5,8)$ because it is length $17$ along $x$, width $5$ along $y$, and height $8$ along $z$. 4. **Vector $\overrightarrow{AH}$:** $$\overrightarrow{AH} = (17-0, 5-0, 8-0) = (17, 5, 8)$$ 5. **Normal vector to plane $EFGH$:** Since $EFGH$ is the base plane parallel to $xy$-plane at height $8$, its normal vector is vertical: $$\vec{n} = (0,0,1)$$ 6. **Angle between vector and plane:** The angle $\theta$ between vector $\overrightarrow{AH}$ and plane $EFGH$ is complementary to the angle $\phi$ between $\overrightarrow{AH}$ and the normal vector $\vec{n}$: $$\theta = 90^\circ - \phi$$ 7. **Calculate $\phi$ using dot product:** $$\cos \phi = \frac{\overrightarrow{AH} \cdot \vec{n}}{|\overrightarrow{AH}| |\vec{n}|} = \frac{8}{\sqrt{17^2 + 5^2 + 8^2} \times 1} = \frac{8}{\sqrt{289 + 25 + 64}} = \frac{8}{\sqrt{378}}$$ 8. **Simplify:** $$\sqrt{378} = \sqrt{9 \times 42} = 3 \sqrt{42}$$ So, $$\cos \phi = \frac{8}{3 \sqrt{42}}$$ 9. **Calculate $\phi$:** $$\phi = \cos^{-1} \left( \frac{8}{3 \sqrt{42}} \right)$$ 10. **Calculate $\theta$:** $$\theta = 90^\circ - \phi$$ 11. **Numerical evaluation:** $$\frac{8}{3 \sqrt{42}} \approx \frac{8}{3 \times 6.4807} = \frac{8}{19.442} \approx 0.4114$$ $$\phi = \cos^{-1}(0.4114) \approx 65.7^\circ$$ $$\theta = 90^\circ - 65.7^\circ = 24.3^\circ$$ **Final answer:** The angle $AH$ makes with the plane $EFGH$ is approximately **24.3 degrees** to 1 decimal place.