Angle Bec Bd1C49

1. **Problem Statement:** We need to find the size of the angle $BEC$ in the cuboid. 2. **Understanding the cuboid and points:** The cuboid has vertices $A$, $B$, $C$, $D$ on the top face and $E$, $F$, $G$, $H$ on the bottom face. 3. **Given dimensions:** - Length $AB = 23$ cm - Width $BC = 15$ cm - Height $AE = 6$ cm 4. **Identify points $B$, $E$, and $C$ in 3D coordinates:** - Let $E$ be at the origin $(0,0,0)$. - Since $AE = 6$ cm is height, $A$ is at $(0,0,6)$. - $B$ is along length from $A$, so $B$ is at $(23,0,6)$. - $C$ is along width from $B$, so $C$ is at $(23,15,6)$. 5. **Vectors for angle $BEC$:** - Vector $EB = B - E = (23,0,6) - (0,0,0) = (23,0,6)$ - Vector $EC = C - E = (23,15,6) - (0,0,0) = (23,15,6)$ 6. **Formula for angle between two vectors:** $$\cos(\theta) = \frac{\vec{EB} \cdot \vec{EC}}{|\vec{EB}| |\vec{EC}|}$$ 7. **Calculate dot product:** $$\vec{EB} \cdot \vec{EC} = 23 \times 23 + 0 \times 15 + 6 \times 6 = 529 + 0 + 36 = 565$$ 8. **Calculate magnitudes:** $$|\vec{EB}| = \sqrt{23^2 + 0^2 + 6^2} = \sqrt{529 + 0 + 36} = \sqrt{565}$$ $$|\vec{EC}| = \sqrt{23^2 + 15^2 + 6^2} = \sqrt{529 + 225 + 36} = \sqrt{790}$$ 9. **Calculate cosine of angle:** $$\cos(\theta) = \frac{565}{\sqrt{565} \times \sqrt{790}} = \frac{565}{\sqrt{565 \times 790}}$$ 10. **Simplify denominator:** $$\sqrt{565 \times 790} = \sqrt{446350} \approx 668.22$$ 11. **Calculate cosine:** $$\cos(\theta) = \frac{565}{668.22} \approx 0.8457$$ 12. **Calculate angle:** $$\theta = \cos^{-1}(0.8457) \approx 32.18^\circ$$ 13. **Final answer:** The size of angle $BEC$ is approximately $32.18^\circ$ to 2 decimal places.