1. The problem is to calculate the angle between the vectors $\vec{a} = (1, 2)$ and $\vec{c} = (-2, 5)$. 2. The formula to find the angle $\theta$ between two vectors $\vec{u}$ and $\vec{v}$ is: $$\cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|}$$ where $\vec{u} \cdot \vec{v}$ is the dot product and $|\vec{u}|$, $|\vec{v}|$ are the magnitudes of the vectors. 3. Calculate the dot product: $$\vec{a} \cdot \vec{c} = (1)(-2) + (2)(5) = -2 + 10 = 8$$ 4. Calculate the magnitudes: $$|\vec{a}| = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}$$ $$|\vec{c}| = \sqrt{(-2)^2 + 5^2} = \sqrt{4 + 25} = \sqrt{29}$$ 5. Substitute into the cosine formula: $$\cos \theta = \frac{8}{\sqrt{5} \times \sqrt{29}} = \frac{8}{\sqrt{145}}$$ 6. Calculate $\theta$ by taking the inverse cosine: $$\theta = \cos^{-1} \left( \frac{8}{\sqrt{145}} \right)$$ 7. Numerically, $$\frac{8}{\sqrt{145}} \approx \frac{8}{12.0416} \approx 0.6644$$ $$\theta \approx \cos^{-1}(0.6644) \approx 48.5^\circ$$ **Final answer:** The angle between $\vec{a}$ and $\vec{c}$ is approximately **48.5 degrees**.