1. **Problem statement:** We have a pyramid with rectangular base ABCD where $AB=20$ cm, $BC=16$ cm, and the vertex $E$ such that $AE=BE=CE=DE=17$ cm. We need to find the angle between the edge $BE$ and the base plane $ABCD$. 2. **Understanding the problem:** The base $ABCD$ is a rectangle, and $E$ is a point above the base such that all edges from $E$ to the vertices of the base are equal in length (17 cm). The angle between edge $BE$ and the base is the angle between the line segment $BE$ and the plane $ABCD$. 3. **Approach:** The angle between a line and a plane is complementary to the angle between the line and the normal vector to the plane. So, if $\theta$ is the angle between $BE$ and the base, then $\theta = 90^\circ - \alpha$, where $\alpha$ is the angle between $BE$ and the normal vector to the base. 4. **Step 1: Set coordinate system.** Place the rectangle $ABCD$ in the $xy$-plane: - Let $A=(0,0,0)$ - $B=(20,0,0)$ since $AB=20$ - $C=(20,16,0)$ since $BC=16$ - $D=(0,16,0)$ 5. **Step 2: Find coordinates of $E$.** Since $AE=BE=CE=DE=17$, $E$ is equidistant from all four vertices. Let $E=(x,y,z)$. 6. **Step 3: Use distance formulas:** - $AE=17 \Rightarrow \sqrt{(x-0)^2+(y-0)^2+(z-0)^2}=17 \Rightarrow x^2+y^2+z^2=289$ - $BE=17 \Rightarrow \sqrt{(x-20)^2+y^2+z^2}=17 \Rightarrow (x-20)^2+y^2+z^2=289$ 7. **Step 4: Subtract equations:** $$x^2+y^2+z^2 = (x-20)^2 + y^2 + z^2$$ Simplify: $$x^2 = (x-20)^2 = x^2 - 40x + 400$$ So, $$x^2 = x^2 - 40x + 400 \Rightarrow 0 = -40x + 400 \Rightarrow 40x = 400 \Rightarrow x=10$$ 8. **Step 5: Use $CE=17$:** $$\sqrt{(x-20)^2 + (y-16)^2 + z^2} = 17$$ Square both sides: $$(x-20)^2 + (y-16)^2 + z^2 = 289$$ Substitute $x=10$: $$(10-20)^2 + (y-16)^2 + z^2 = 289$$ $$100 + (y-16)^2 + z^2 = 289$$ $$(y-16)^2 + z^2 = 189$$ 9. **Step 6: Use $DE=17$:** $$\sqrt{(x-0)^2 + (y-16)^2 + z^2} = 17$$ Square both sides: $$x^2 + (y-16)^2 + z^2 = 289$$ Substitute $x=10$: $$100 + (y-16)^2 + z^2 = 289$$ $$(y-16)^2 + z^2 = 189$$ 10. **Step 7: Use $AE=17$ again:** $$x^2 + y^2 + z^2 = 289$$ Substitute $x=10$: $$100 + y^2 + z^2 = 289$$ $$y^2 + z^2 = 189$$ 11. **Step 8: From steps 8 and 9:** $$(y-16)^2 + z^2 = 189$$ $$y^2 + z^2 = 189$$ Subtract second from first: $$(y-16)^2 - y^2 = 0$$ Expand: $$y^2 - 32y + 256 - y^2 = 0 \Rightarrow -32y + 256 = 0 \Rightarrow 32y = 256 \Rightarrow y=8$$ 12. **Step 9: Find $z$:** From $y^2 + z^2 = 189$: $$8^2 + z^2 = 189 \Rightarrow 64 + z^2 = 189 \Rightarrow z^2 = 125 \Rightarrow z = \sqrt{125} = 5\sqrt{5}$$ 13. **Step 10: Coordinates of $E$ are:** $$E = (10, 8, 5\sqrt{5})$$ 14. **Step 11: Vector $\overrightarrow{BE}$:** $$B = (20,0,0)$$ $$\overrightarrow{BE} = E - B = (10-20, 8-0, 5\sqrt{5} - 0) = (-10, 8, 5\sqrt{5})$$ 15. **Step 12: Normal vector to base $ABCD$:** Since base is in $xy$-plane, normal vector is $$\vec{n} = (0,0,1)$$ 16. **Step 13: Find angle $\alpha$ between $\overrightarrow{BE}$ and $\vec{n}$:** Use dot product formula: $$\cos \alpha = \frac{\overrightarrow{BE} \cdot \vec{n}}{|\overrightarrow{BE}| |\vec{n}|}$$ Calculate dot product: $$\overrightarrow{BE} \cdot \vec{n} = (-10)(0) + 8(0) + 5\sqrt{5}(1) = 5\sqrt{5}$$ Magnitude of $\overrightarrow{BE}$: $$|\overrightarrow{BE}| = \sqrt{(-10)^2 + 8^2 + (5\sqrt{5})^2} = \sqrt{100 + 64 + 125} = \sqrt{289} = 17$$ Magnitude of $\vec{n}$ is 1. So, $$\cos \alpha = \frac{5\sqrt{5}}{17}$$ 17. **Step 14: Calculate $\alpha$:** $$\alpha = \cos^{-1}\left(\frac{5\sqrt{5}}{17}\right)$$ 18. **Step 15: Find angle $\theta$ between $BE$ and base:** $$\theta = 90^\circ - \alpha = 90^\circ - \cos^{-1}\left(\frac{5\sqrt{5}}{17}\right)$$ **Final answer:** $$\boxed{\theta = 90^\circ - \cos^{-1}\left(\frac{5\sqrt{5}}{17}\right)}$$ This is the angle between edge $BE$ and the base $ABCD$.