1. **Calculate angle EAC in the first pyramid:** The base ABCD is a square with side $AB=5$ cm. Point $E$ is 10 cm above the base vertically. 2. Find the diagonal $AC$ of the square base: $$AC=\sqrt{AB^2+BC^2}=\sqrt{5^2+5^2}=\sqrt{50}=5\sqrt{2}\text{ cm}$$ 3. Coordinates (assuming $A$ at origin): $A=(0,0,0)$, $C=(5,5,0)$, $E=(2.5,2.5,10)$ since $E$ is above midpoint of $AC$. 4. Calculate vectors: $$\overrightarrow{AE}=(2.5,2.5,10),\quad \overrightarrow{AC}=(5,5,0)$$ 5. Use dot product formula for angle between $\overrightarrow{AE}$ and $\overrightarrow{AC}$: $$\cos\theta=\frac{\overrightarrow{AE}\cdot\overrightarrow{AC}}{|AE||AC|}$$ Dot product: $$\overrightarrow{AE}\cdot\overrightarrow{AC} = 2.5\times 5 + 2.5\times 5 + 10\times 0 = 25$$ Magnitudes: $$|AE|=\sqrt{2.5^2+2.5^2+10^2}=\sqrt{6.25+6.25+100}=\sqrt{112.5}\approx 10.6066$$ $$|AC|=5\sqrt{2}\approx 7.0711$$ Calculate cosine: $$\cos\theta=\frac{25}{10.6066\times 7.0711}\approx 0.3333$$ Find angle: $$\theta=\cos^{-1}(0.3333)\approx 70.53^\circ$$ --- 6. **Calculate the angle the line AF makes with the plane ABCD in the triangular prism:** Given triangle $ACD$ with $CD=20$ cm, $AD=30$ cm, and angle $FDC=35^\circ$. Points $E,F$ are vertically above $A,D$ respectively. 7. Find length $AC$ via Law of Cosines in $\triangle ACD$: $$AC^2=AD^2+CD^2-2\times AD \times CD \times \cos 35^\circ$$ $$AC^2=30^2+20^2-2\times 30 \times 20 \times \cos 35^\circ=900+400-1200 \times 0.8192=1300 - 983.04=316.96$$ $$AC=\sqrt{316.96} \approx 17.8\text{ cm}$$ 8. Plane $ABCD$ is horizontal; line $AF$ is the segment between $A$ and $F$, where $F$ is vertically above $D$. 9. Vector $\overrightarrow{AF}$ is vertical rise from $A$ to $F$ involving height equal to length of prism (not given, so assume vertical length not specified, we define essentially the height). Actually, angle $FDC=35^\circ$ is the angle between edge $FD$ and $DC$. From the geometry of prism, the angle between $AF$ and plane $ABCD$ is equal to complement of angle $FDC$ since $F$ is vertically above $D$. So angle with plane: $$90^\circ - 35^\circ = 55^\circ$$ Because line $AF$ is vertical and plane is horizontal, angle between line $AF$ and plane $ABCD$ is $55^\circ$. (Note: This assumes vertical prism edges. If further calculations needed, please specify.) --- 10. **Calculate volume of the second pyramid:** Base ABCD is square with side $AB=15$ cm and apex $P$ above base with angle $PAC=65^\circ$. 11. Find diagonal $AC$ of base: $$AC = \sqrt{15^2 + 15^2} = \sqrt{450} = 15\sqrt{2} \approx 21.213\text{ cm}$$ 12. Triangle $PAC$ includes angle $65^\circ$ at $A$. Length $PA$ (height) is unknown; find it using angle definition. Let height $h=PA$ and triangle formed by $P,A,C$. 13. Using angle $PAC=65^\circ$ and base $AC$ calculate height $h$ based on projection. From triangle $PAC$, projection of $PA$ onto the base plane forms angle $65^\circ$ with diagonal $AC$. So height $h = AC \times \tan 65^\circ = 21.213 \times 2.1445 \approx 45.45$ cm 14. Volume $V$ of pyramid: $$V=\frac{1}{3} \times \text{Area of base} \times \text{height}$$ Base area: $$15 \times 15 = 225 \text{ cm}^2$$ Volume: $$V = \frac{1}{3} \times 225 \times 45.45 \approx 3408.75 \text{ cm}^3$$ --- **Final answers:** - Angle $EAC \approx 70.5^\circ$ - Angle between line $AF$ and plane $ABCD = 55^\circ$ - Volume of pyramid $\approx 3409$ cm$^3$ (to 4 significant figures)