1. **State the problem:** Find the angle between the two lines given by the parametric equations: Line 1: $x = 9 - 6t$, $y = -5t - 10$, $z = 0$ Line 2: $x = -5t - 7$, $y = 2t$, $z = -4t - 5$ 2. **Formula used:** The angle $\theta$ between two lines in 3D can be found using the direction vectors of the lines. If $\mathbf{v_1}$ and $\mathbf{v_2}$ are direction vectors, then $$\cos \theta = \frac{|\mathbf{v_1} \cdot \mathbf{v_2}|}{\|\mathbf{v_1}\| \|\mathbf{v_2}\|}$$ where $\cdot$ is the dot product and $\|\cdot\|$ is the vector magnitude. 3. **Find direction vectors:** From Line 1, direction vector $\mathbf{v_1} = \langle -6, -5, 0 \rangle$ From Line 2, direction vector $\mathbf{v_2} = \langle -5, 2, -4 \rangle$ 4. **Calculate dot product:** $$\mathbf{v_1} \cdot \mathbf{v_2} = (-6)(-5) + (-5)(2) + (0)(-4) = 30 - 10 + 0 = 20$$ 5. **Calculate magnitudes:** $$\|\mathbf{v_1}\| = \sqrt{(-6)^2 + (-5)^2 + 0^2} = \sqrt{36 + 25} = \sqrt{61}$$ $$\|\mathbf{v_2}\| = \sqrt{(-5)^2 + 2^2 + (-4)^2} = \sqrt{25 + 4 + 16} = \sqrt{45}$$ 6. **Calculate cosine of angle:** $$\cos \theta = \frac{|20|}{\sqrt{61} \times \sqrt{45}} = \frac{20}{\sqrt{2745}}$$ 7. **Calculate angle $\theta$:** $$\theta = \cos^{-1} \left( \frac{20}{\sqrt{2745}} \right)$$ Calculate numeric value: $$\sqrt{2745} \approx 52.38$$ $$\cos \theta \approx \frac{20}{52.38} \approx 0.3819$$ $$\theta \approx \cos^{-1}(0.3819) \approx 67.5^\circ$$ 8. **Final answer:** The angle between the two lines is approximately **68 degrees** (to the nearest degree).