1. **Stating the problem:** We are given a 3D shape with an angle $\theta$ at the left-front bottom vertex formed between a slanted segment and a vertical dashed line inside the prism. The lengths given are 8 cm, 6 cm, and 4 cm, which correspond to edges of the prism. We need to find the value of the angle $\theta$. 2. **Understanding the geometry:** The angle $\theta$ is between the slanted segment (from the bottom vertex to the apex) and the vertical dashed line inside the prism. The vertical dashed line is 6 cm, the height. The base edges are 8 cm and 4 cm. 3. **Using the dot product formula for angle between vectors:** The angle between two vectors $\mathbf{u}$ and $\mathbf{v}$ is given by $$\cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|}$$ 4. **Defining vectors:** Let the vertical vector be $\mathbf{v} = (0,6,0)$ (height along y-axis). The slanted segment vector $\mathbf{u}$ goes from the bottom vertex to the apex. The apex is at $(8,6,4)$ relative to the bottom vertex at $(0,0,0)$, so $$\mathbf{u} = (8,6,4)$$ 5. **Calculate dot product:** $$\mathbf{u} \cdot \mathbf{v} = 8 \times 0 + 6 \times 6 + 4 \times 0 = 36$$ 6. **Calculate magnitudes:** $$\|\mathbf{u}\| = \sqrt{8^2 + 6^2 + 4^2} = \sqrt{64 + 36 + 16} = \sqrt{116}$$ $$\|\mathbf{v}\| = \sqrt{0^2 + 6^2 + 0^2} = 6$$ 7. **Calculate cosine of $\theta$:** $$\cos \theta = \frac{36}{6 \times \sqrt{116}} = \frac{36}{6\sqrt{116}} = \frac{6}{\sqrt{116}}$$ 8. **Simplify $\sqrt{116}$:** $$\sqrt{116} = \sqrt{4 \times 29} = 2\sqrt{29}$$ 9. **Substitute back:** $$\cos \theta = \frac{6}{2\sqrt{29}} = \frac{3}{\sqrt{29}}$$ 10. **Calculate $\theta$:** $$\theta = \cos^{-1} \left( \frac{3}{\sqrt{29}} \right) \approx \cos^{-1}(0.557) \approx 56.1^\circ$$ **Final answer:** $$\boxed{\theta \approx 56.1^\circ}$$