Pipe Angle Bff220

1. **Problem Statement:** We are given a 3D pipe diagram with points $A$, $B$, $C$, $D$, and $F$. The lengths $a = 6$ ft and $h = 1.5$ ft are provided. We need to find the angle $\theta$ between the pipe segments $BA$ and $BC$ at point $B$. 2. **Understanding the Geometry:** - Segment $BA$ goes from $B$ to $A$. - Segment $BC$ goes from $B$ to $C$. We want the angle between vectors $\overrightarrow{BA}$ and $\overrightarrow{BC}$. 3. **Vectors Setup:** Assuming the coordinate system based on the figure description: - Let $B$ be at the origin $(0,0,0)$. - Point $A$ is located at $(a,0,h) = (6,0,1.5)$. - Point $C$ is located at $(0,4,0)$ (since the horizontal dashed line from $C$ to $D$ is along the y-axis with length 4 ft). 4. **Calculate vectors:** $$\overrightarrow{BA} = A - B = (6,0,1.5) - (0,0,0) = (6,0,1.5)$$ $$\overrightarrow{BC} = C - B = (0,4,0) - (0,0,0) = (0,4,0)$$ 5. **Formula for angle between two vectors:** $$\cos \theta = \frac{\overrightarrow{BA} \cdot \overrightarrow{BC}}{|\overrightarrow{BA}| |\overrightarrow{BC}|}$$ where $\cdot$ is the dot product and $|\cdot|$ is the magnitude. 6. **Calculate dot product:** $$\overrightarrow{BA} \cdot \overrightarrow{BC} = (6)(0) + (0)(4) + (1.5)(0) = 0$$ 7. **Calculate magnitudes:** $$|\overrightarrow{BA}| = \sqrt{6^2 + 0^2 + 1.5^2} = \sqrt{36 + 0 + 2.25} = \sqrt{38.25}$$ $$|\overrightarrow{BC}| = \sqrt{0^2 + 4^2 + 0^2} = \sqrt{16} = 4$$ 8. **Calculate $\cos \theta$:** $$\cos \theta = \frac{0}{\sqrt{38.25} \times 4} = 0$$ 9. **Calculate $\theta$:** $$\theta = \cos^{-1}(0) = 90^\circ$$ 10. **Final answer:** The angle $\theta$ between pipe segments $BA$ and $BC$ is $$\boxed{90.0^\circ}$$ This is to three significant figures.