1. Stating the problem: We have a pyramid ABCS with base ABC being a right triangle with legs $|AB|=6$ and $|AC|=8$. The lateral edges $|AS|=|BS|=|CS|=10$. We need to find the volume of the pyramid and the sine of the angle between the lateral face ACS and the base ABC. 2. Find the area of the base triangle ABC. Since ABC is right-angled at A, area is $$\text{Area}_{ABC} = \frac{1}{2} \times |AB| \times |AC| = \frac{1}{2} \times 6 \times 8 = 24.$$ 3. Find the height of the pyramid (distance from S to the plane ABC). Since $|AS|=|BS|=|CS|=10$, point S is equidistant from A, B, and C. 4. Calculate the length of BC using Pythagoras: $$|BC| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10.$$ 5. Since S is equidistant from A, B, and C, and $|AS|=|BS|=|CS|=10$, S lies on the sphere centered at A with radius 10, and also on spheres centered at B and C with radius 10. 6. To find the height $h$ of the pyramid, drop a perpendicular from S to the plane ABC. Let $H$ be the foot of the perpendicular. 7. Coordinates setup for convenience: - Place A at origin $(0,0,0)$. - Place B at $(6,0,0)$. - Place C at $(0,8,0)$. 8. Let $S = (x,y,z)$ with $z=h$ (height above the plane). 9. Using distances: $$|AS|^2 = x^2 + y^2 + z^2 = 10^2 = 100,$$ $$|BS|^2 = (x-6)^2 + y^2 + z^2 = 100,$$ $$|CS|^2 = x^2 + (y-8)^2 + z^2 = 100.$$ 10. Subtract $|AS|^2$ from $|BS|^2$: $$(x-6)^2 + y^2 + z^2 - (x^2 + y^2 + z^2) = 0 \Rightarrow (x-6)^2 - x^2 = 0,$$ $$x^2 - 12x + 36 - x^2 = 0 \Rightarrow -12x + 36 = 0 \Rightarrow x = 3.$$ 11. Subtract $|AS|^2$ from $|CS|^2$: $$x^2 + (y-8)^2 + z^2 - (x^2 + y^2 + z^2) = 0 \Rightarrow (y-8)^2 - y^2 = 0,$$ $$y^2 - 16y + 64 - y^2 = 0 \Rightarrow -16y + 64 = 0 \Rightarrow y = 4.$$ 12. Substitute $x=3$, $y=4$ into $|AS|^2$: $$3^2 + 4^2 + z^2 = 100 \Rightarrow 9 + 16 + z^2 = 100 \Rightarrow z^2 = 75 \Rightarrow z = \sqrt{75} = 5\sqrt{3}.$$ 13. Height of the pyramid is $h = 5\sqrt{3}$. 14. Volume of the pyramid: $$V = \frac{1}{3} \times \text{Area}_{ABC} \times h = \frac{1}{3} \times 24 \times 5\sqrt{3} = 40\sqrt{3}.$$ 15. To find $\sin$ of the angle between face ACS and base ABC, find the angle between planes ABC and ACS. 16. Normal vector to base ABC is along $z$-axis: $$\vec{n}_{ABC} = (0,0,1).$$ 17. Find normal vector to face ACS. 18. Vectors in face ACS: $$\vec{AC} = (0-0,8-0,0-0) = (0,8,0),$$ $$\vec{AS} = (3-0,4-0,5\sqrt{3}-0) = (3,4,5\sqrt{3}).$$ 19. Normal vector to ACS is cross product: $$\vec{n}_{ACS} = \vec{AC} \times \vec{AS} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 8 & 0 \\ 3 & 4 & 5\sqrt{3} \end{vmatrix} = (8 \times 5\sqrt{3} - 0, 0 - 0, 0 - 24) = (40\sqrt{3}, 0, -24).$$ 20. Compute magnitudes: $$|\vec{n}_{ACS}| = \sqrt{(40\sqrt{3})^2 + 0^2 + (-24)^2} = \sqrt{4800 + 576} = \sqrt{5376} = 24\sqrt{14},$$ $$|\vec{n}_{ABC}| = 1.$$ 21. Cosine of angle between planes is: $$\cos \theta = \frac{|\vec{n}_{ABC} \cdot \vec{n}_{ACS}|}{|\vec{n}_{ABC}||\vec{n}_{ACS}|} = \frac{|0 \times 40\sqrt{3} + 0 \times 0 + 1 \times (-24)|}{1 \times 24\sqrt{14}} = \frac{24}{24\sqrt{14}} = \frac{1}{\sqrt{14}}.$$ 22. Sine of angle between planes is: $$\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \frac{1}{14}} = \sqrt{\frac{13}{14}} = \frac{\sqrt{13}}{\sqrt{14}}.$$ Final answers: - Volume of pyramid: $40\sqrt{3}$ - $\sin$ of angle between face ACS and base ABC: $\frac{\sqrt{13}}{\sqrt{14}}$