1. **Problem statement:** Find the area, sine, and cosine of each angle of the triangle with vertices A(0,0,0), B(4,-1,3), and C(1,2,3). 2. **Step 1: Find vectors AB and AC.** $$\vec{AB} = B - A = (4 - 0, -1 - 0, 3 - 0) = (4, -1, 3)$$ $$\vec{AC} = C - A = (1 - 0, 2 - 0, 3 - 0) = (1, 2, 3)$$ 3. **Step 2: Calculate the lengths of AB and AC.** $$|\vec{AB}| = \sqrt{4^2 + (-1)^2 + 3^2} = \sqrt{16 + 1 + 9} = \sqrt{26}$$ $$|\vec{AC}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$$ 4. **Step 3: Find the vector BC.** $$\vec{BC} = C - B = (1 - 4, 2 - (-1), 3 - 3) = (-3, 3, 0)$$ 5. **Step 4: Calculate the length of BC.** $$|\vec{BC}| = \sqrt{(-3)^2 + 3^2 + 0^2} = \sqrt{9 + 9 + 0} = \sqrt{18} = 3\sqrt{2}$$ 6. **Step 5: Calculate the area of the triangle using the cross product of AB and AC.** $$\vec{AB} \times \vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -1 & 3 \\ 1 & 2 & 3 \end{vmatrix} = \mathbf{i}((-1)(3) - 3(2)) - \mathbf{j}(4(3) - 3(1)) + \mathbf{k}(4(2) - (-1)(1))$$ $$= \mathbf{i}(-3 - 6) - \mathbf{j}(12 - 3) + \mathbf{k}(8 + 1) = -9\mathbf{i} - 9\mathbf{j} + 9\mathbf{k}$$ 7. **Step 6: Find the magnitude of the cross product.** $$|\vec{AB} \times \vec{AC}| = \sqrt{(-9)^2 + (-9)^2 + 9^2} = \sqrt{81 + 81 + 81} = \sqrt{243} = 9\sqrt{3}$$ 8. **Step 7: Calculate the area of the triangle.** $$\text{Area} = \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{1}{2} \times 9\sqrt{3} = \frac{9\sqrt{3}}{2}$$ 9. **Step 8: Calculate the cosines of the angles using dot products.** - Angle at A between AB and AC: $$\cos \theta_A = \frac{\vec{AB} \cdot \vec{AC}}{|\vec{AB}||\vec{AC}|}$$ $$\vec{AB} \cdot \vec{AC} = 4 \times 1 + (-1) \times 2 + 3 \times 3 = 4 - 2 + 9 = 11$$ $$\cos \theta_A = \frac{11}{\sqrt{26} \times \sqrt{14}} = \frac{11}{\sqrt{364}} = \frac{11}{2\sqrt{91}}$$ - Angle at B between BA and BC (note: BA = -AB): $$\vec{BA} = -\vec{AB} = (-4, 1, -3)$$ $$\vec{BA} \cdot \vec{BC} = (-4)(-3) + 1 \times 3 + (-3) \times 0 = 12 + 3 + 0 = 15$$ $$|\vec{BA}| = |\vec{AB}| = \sqrt{26}, \quad |\vec{BC}| = 3\sqrt{2}$$ $$\cos \theta_B = \frac{15}{\sqrt{26} \times 3\sqrt{2}} = \frac{15}{3\sqrt{52}} = \frac{5}{\sqrt{52}} = \frac{5}{2\sqrt{13}}$$ - Angle at C between CA and CB (CA = -AC, CB = -BC): $$\vec{CA} = -\vec{AC} = (-1, -2, -3), \quad \vec{CB} = -\vec{BC} = (3, -3, 0)$$ $$\vec{CA} \cdot \vec{CB} = (-1)(3) + (-2)(-3) + (-3)(0) = -3 + 6 + 0 = 3$$ $$|\vec{CA}| = |\vec{AC}| = \sqrt{14}, \quad |\vec{CB}| = |\vec{BC}| = 3\sqrt{2}$$ $$\cos \theta_C = \frac{3}{\sqrt{14} \times 3\sqrt{2}} = \frac{3}{3\sqrt{28}} = \frac{1}{\sqrt{28}} = \frac{1}{2\sqrt{7}}$$ 10. **Step 9: Calculate the sines of the angles using** $$\sin \theta = \sqrt{1 - \cos^2 \theta}$$ - For angle A: $$\sin \theta_A = \sqrt{1 - \left(\frac{11}{2\sqrt{91}}\right)^2} = \sqrt{1 - \frac{121}{4 \times 91}} = \sqrt{1 - \frac{121}{364}} = \sqrt{\frac{243}{364}} = \frac{9\sqrt{3}}{2\sqrt{91}}$$ - For angle B: $$\sin \theta_B = \sqrt{1 - \left(\frac{5}{2\sqrt{13}}\right)^2} = \sqrt{1 - \frac{25}{4 \times 13}} = \sqrt{1 - \frac{25}{52}} = \sqrt{\frac{27}{52}} = \frac{3\sqrt{3}}{2\sqrt{13}}$$ - For angle C: $$\sin \theta_C = \sqrt{1 - \left(\frac{1}{2\sqrt{7}}\right)^2} = \sqrt{1 - \frac{1}{4 \times 7}} = \sqrt{1 - \frac{1}{28}} = \sqrt{\frac{27}{28}} = \frac{3\sqrt{3}}{2\sqrt{7}}$$ **Final answers:** - Area of triangle = $\frac{9\sqrt{3}}{2}$ - Cosines of angles: $\cos \theta_A = \frac{11}{2\sqrt{91}}$, $\cos \theta_B = \frac{5}{2\sqrt{13}}$, $\cos \theta_C = \frac{1}{2\sqrt{7}}$ - Sines of angles: $\sin \theta_A = \frac{9\sqrt{3}}{2\sqrt{91}}$, $\sin \theta_B = \frac{3\sqrt{3}}{2\sqrt{13}}$, $\sin \theta_C = \frac{3\sqrt{3}}{2\sqrt{7}}$