1. **Problem statement:** Find the angle between the line segments AB and AC where $A=(2,3)$, $B=(-1,-4)$, and $C=(0,-2)$. 2. **Formula used:** The angle $\theta$ between two vectors $\vec{u}$ and $\vec{v}$ is given by $$\cos\theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|}$$ where $\vec{u} \cdot \vec{v}$ is the dot product and $|\vec{u}|, |\vec{v}|$ are the magnitudes. 3. **Find vectors:** $\vec{AB} = B - A = (-1 - 2, -4 - 3) = (-3, -7)$ $\vec{AC} = C - A = (0 - 2, -2 - 3) = (-2, -5)$ 4. **Calculate dot product:** $$\vec{AB} \cdot \vec{AC} = (-3)(-2) + (-7)(-5) = 6 + 35 = 41$$ 5. **Calculate magnitudes:** $$|\vec{AB}| = \sqrt{(-3)^2 + (-7)^2} = \sqrt{9 + 49} = \sqrt{58}$$ $$|\vec{AC}| = \sqrt{(-2)^2 + (-5)^2} = \sqrt{4 + 25} = \sqrt{29}$$ 6. **Calculate cosine of angle:** $$\cos\theta = \frac{41}{\sqrt{58} \times \sqrt{29}} = \frac{41}{\sqrt{1682}}$$ 7. **Calculate angle $\theta$:** $$\theta = \cos^{-1}\left(\frac{41}{\sqrt{1682}}\right)$$ This is the exact angle between the line segments AB and AC. **Final answer:** $$\boxed{\theta = \cos^{-1}\left(\frac{41}{\sqrt{1682}}\right)}$$