1. **Problem Statement:** Given a cube with edges of length 6 cm, find: a) Length BD b) Length AS c) Length BS d) Angle SBD e) Angle ASB 2. **Cube Properties and Coordinates:** Assume the cube vertices are positioned in 3D space with A at the origin $(0,0,0)$. Edges AB, BC, AD, and DR are each 6 cm. Assign coordinates: - $A = (0,0,0)$ - $B = (6,0,0)$ - $D = (0,6,0)$ - $S = (6,6,6)$ (top far corner) 3. **Find BD:** $BD$ is the diagonal of the square face ABD. Using distance formula: $$BD = \sqrt{(0-6)^2 + (6-0)^2 + (0-0)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}$$ 4. **Find AS:** $AS$ is the space diagonal of the cube from $(0,0,0)$ to $(6,6,6)$. $$AS = \sqrt{(6-0)^2 + (6-0)^2 + (6-0)^2} = \sqrt{36 + 36 + 36} = \sqrt{108} = 6\sqrt{3}$$ 5. **Find BS:** $B = (6,0,0)$ and $S = (6,6,6)$. $$BS = \sqrt{(6-6)^2 + (6-0)^2 + (6-0)^2} = \sqrt{0 + 36 + 36} = \sqrt{72} = 6\sqrt{2}$$ 6. **Find angle SBD:** Vectors: $\overrightarrow{BD} = D - B = (0-6,6-0,0-0) = (-6,6,0)$ $\overrightarrow{BS} = S - B = (6-6,6-0,6-0) = (0,6,6)$ Dot product: $$\overrightarrow{BD} \cdot \overrightarrow{BS} = (-6)(0) + 6(6) + 0(6) = 36$$ Magnitudes: $$|\overrightarrow{BD}| = 6\sqrt{2}, \quad |\overrightarrow{BS}| = 6\sqrt{2}$$ Angle $\theta$ between vectors: $$\cos \theta = \frac{\overrightarrow{BD} \cdot \overrightarrow{BS}}{|\overrightarrow{BD}||\overrightarrow{BS}|} = \frac{36}{6\sqrt{2} \times 6\sqrt{2}} = \frac{36}{72} = \frac{1}{2}$$ $$\theta = \cos^{-1}(\frac{1}{2}) = 60^\circ$$ 7. **Find angle ASB:** Vectors: $\overrightarrow{AS} = S - A = (6,6,6)$ $\overrightarrow{BS} = S - B = (0,6,6)$ Dot product: $$\overrightarrow{AS} \cdot \overrightarrow{BS} = 6(0) + 6(6) + 6(6) = 0 + 36 + 36 = 72$$ Magnitudes: $$|\overrightarrow{AS}| = 6\sqrt{3}, \quad |\overrightarrow{BS}| = 6\sqrt{2}$$ Angle $\phi$: $$\cos \phi = \frac{72}{6\sqrt{3} \times 6\sqrt{2}} = \frac{72}{36\sqrt{6}} = \frac{2}{\sqrt{6}} = \frac{2\sqrt{6}}{6} = \frac{\sqrt{6}}{3}$$ $$\phi = \cos^{-1}(\frac{\sqrt{6}}{3}) \approx 35.26^\circ$$ **Final answers:** - $BD = 6\sqrt{2}$ cm - $AS = 6\sqrt{3}$ cm - $BS = 6\sqrt{2}$ cm - $\angle SBD = 60^\circ$ - $\angle ASB \approx 35.26^\circ$