1. **State the problem:** We need to find the size of angle $\angle CME$ in the triangular prism $ABCDEF$ where $M$ is the midpoint of $AC$, $AC=40$ cm, $\angle CAE=35^\circ$, and $\angle ABF=90^\circ$. The prism has a horizontal rectangular base $ABCD$. 2. **Understand the geometry:** - $M$ is midpoint of $AC$, so $AM=MC=\frac{40}{2}=20$ cm. - $\angle ABF=90^\circ$ tells us that $AB$ is perpendicular to $BF$. - $\angle CAE=35^\circ$ is an angle involving points $C$, $A$, and $E$. 3. **Identify vectors for points:** - Let’s place the prism in a coordinate system for clarity. - Place $A$ at origin $(0,0,0)$. - Since $ABCD$ is a rectangle and $AC=40$ cm, let $C$ be at $(40,0,0)$. - $M$ is midpoint of $AC$ at $(20,0,0)$. 4. **Find coordinates of $E$:** - $E$ is vertically above $D$. - Since $ABCD$ is rectangle, $D$ is at $(0,y,0)$ for some $y$. - $E$ is at $(0,y,h)$ where $h$ is height. 5. **Use $\angle CAE=35^\circ$ to find $h$ and $y$:** - Vector $CA = A - C = (0-40,0-0,0-0) = (-40,0,0)$. - Vector $AE = E - A = (0-0,y-0,h-0) = (0,y,h)$. - The angle between vectors $CA$ and $AE$ is $35^\circ$. 6. **Calculate $\cos 35^\circ$ using dot product:** $$\cos 35^\circ = \frac{CA \cdot AE}{|CA||AE|} = \frac{(-40)(0) + 0 \cdot y + 0 \cdot h}{40 \sqrt{y^2 + h^2}} = 0$$ - This implies $\cos 35^\circ = 0$, which is false. 7. **Re-examine vectors:** - The vector $CA$ points from $C$ to $A$, so $CA = A - C = (-40,0,0)$. - The vector $AE$ points from $A$ to $E$, so $AE = (0,y,h)$. 8. **Dot product:** $$CA \cdot AE = (-40)(0) + 0 \cdot y + 0 \cdot h = 0$$ - So the dot product is zero, meaning $CA$ and $AE$ are perpendicular, but given angle is $35^\circ$. 9. **Adjust coordinate assignment:** - Let’s place $A$ at origin $(0,0,0)$. - Let $C$ be at $(40,0,0)$. - Let $D$ be at $(0,d,0)$. - Then $E$ is at $(0,d,h)$. 10. **Vector $CA = C - A = (40,0,0)$, vector $AE = E - A = (0,d,h)$. 11. **Dot product:** $$CA \cdot AE = 40 \times 0 + 0 \times d + 0 \times h = 0$$ - Again zero, so angle between $CA$ and $AE$ is $90^\circ$, contradicting $35^\circ$. 12. **Try $CA = A - C = (-40,0,0)$ and $AE = E - A = (0,d,h)$:** $$CA \cdot AE = (-40)(0) + 0 \times d + 0 \times h = 0$$ - Still zero. 13. **Conclusion:** $CA$ and $AE$ are perpendicular, but problem states $35^\circ$. 14. **Alternative approach:** Assume $AC$ lies along $x$-axis, $AB$ along $y$-axis, and height along $z$-axis. - $A = (0,0,0)$ - $C = (40,0,0)$ - $B = (0,b,0)$ - $D = (40,b,0)$ - $E = (40,b,h)$ 15. **$\angle CAE = 35^\circ$ means angle between vectors $AC$ and $AE$ at $A$:** - $AC = C - A = (40,0,0)$ - $AE = E - A = (40,b,h)$ 16. **Calculate $\cos 35^\circ$:** $$\cos 35^\circ = \frac{AC \cdot AE}{|AC||AE|} = \frac{40 \times 40 + 0 \times b + 0 \times h}{40 \sqrt{40^2 + b^2 + h^2}} = \frac{1600}{40 \sqrt{1600 + b^2 + h^2}} = \frac{40}{\sqrt{1600 + b^2 + h^2}}$$ 17. **Solve for $b^2 + h^2$:** $$\cos 35^\circ = \frac{40}{\sqrt{1600 + b^2 + h^2}} \Rightarrow \sqrt{1600 + b^2 + h^2} = \frac{40}{\cos 35^\circ}$$ $$1600 + b^2 + h^2 = \frac{1600}{\cos^2 35^\circ}$$ $$b^2 + h^2 = \frac{1600}{\cos^2 35^\circ} - 1600$$ 18. **Calculate numeric value:** $$\cos 35^\circ \approx 0.819152$$ $$b^2 + h^2 = \frac{1600}{0.819152^2} - 1600 = \frac{1600}{0.671} - 1600 \approx 2384.5 - 1600 = 784.5$$ 19. **Use $\angle ABF = 90^\circ$ to find relation between $b$ and $h$:** - $F$ is above $B$, so $B = (0,b,0)$, $F = (0,b,h)$. - $AB$ is along $y$-axis: $AB = B - A = (0,b,0)$ - $BF$ is vertical: $F - B = (0,0,h)$ - $AB \perp BF$ confirmed. 20. **Find vector $CM$ and $ME$ to find $\angle CME$:** - $M$ midpoint of $AC$: $M = (20,0,0)$ - $C = (40,0,0)$ - $E = (40,b,h)$ 21. **Vectors:** $$CM = M - C = (20 - 40, 0 - 0, 0 - 0) = (-20, 0, 0)$$ $$ME = E - M = (40 - 20, b - 0, h - 0) = (20, b, h)$$ 22. **Calculate $\cos \angle CME$:** $$\cos \angle CME = \frac{CM \cdot ME}{|CM||ME|} = \frac{(-20)(20) + 0 \times b + 0 \times h}{20 \sqrt{20^2 + b^2 + h^2}} = \frac{-400}{20 \sqrt{400 + b^2 + h^2}} = \frac{-20}{\sqrt{400 + b^2 + h^2}}$$ 23. **Calculate $b^2 + h^2$ from step 18:** $$b^2 + h^2 = 784.5$$ 24. **Calculate denominator:** $$\sqrt{400 + 784.5} = \sqrt{1184.5} \approx 34.42$$ 25. **Calculate $\cos \angle CME$:** $$\cos \angle CME = \frac{-20}{34.42} \approx -0.581$$ 26. **Find angle:** $$\angle CME = \cos^{-1}(-0.581) \approx 125.6^\circ$$ 27. **Final answer:** $$\boxed{125^\circ}$$ (to 3 significant figures)