1. **Problem statement:** Given three non-zero vectors $\vec{A}, \vec{B}, \vec{C}$ such that $\vec{A} + \vec{B} = \vec{C}$, with $\|\vec{A}\| = \|\vec{B}\|$ and $\|\vec{C}\| = \sqrt{3} \|\vec{A}\|$, find the measure of the angle between $\vec{B}$ and $\vec{C}$. 2. **Formula and rules:** The magnitude of the sum of two vectors is given by $$\|\vec{A} + \vec{B}\|^2 = \|\vec{A}\|^2 + \|\vec{B}\|^2 + 2 \|\vec{A}\| \|\vec{B}\| \cos \theta,$$ where $\theta$ is the angle between $\vec{A}$ and $\vec{B}$. 3. **Apply given conditions:** Let $\|\vec{A}\| = \|\vec{B}\| = a$. Then, $$\|\vec{C}\|^2 = \|\vec{A} + \vec{B}\|^2 = a^2 + a^2 + 2a^2 \cos \theta = 2a^2 (1 + \cos \theta).$$ Given $\|\vec{C}\| = \sqrt{3} a$, so $$\|\vec{C}\|^2 = 3a^2.$$ 4. **Equate and solve for $\cos \theta$:** $$3a^2 = 2a^2 (1 + \cos \theta) \implies 3 = 2 (1 + \cos \theta) \implies 1 + \cos \theta = \frac{3}{2} \implies \cos \theta = \frac{1}{2}.$$ 5. **Find $\theta$:** $$\cos \theta = \frac{1}{2} \implies \theta = \frac{\pi}{3}.$$ 6. **Find angle between $\vec{B}$ and $\vec{C}$:** Use the dot product formula: $$\vec{B} \cdot \vec{C} = \|\vec{B}\| \|\vec{C}\| \cos \phi,$$ where $\phi$ is the angle between $\vec{B}$ and $\vec{C}$. Since $\vec{C} = \vec{A} + \vec{B}$, $$\vec{B} \cdot \vec{C} = \vec{B} \cdot (\vec{A} + \vec{B}) = \vec{B} \cdot \vec{A} + \vec{B} \cdot \vec{B} = a^2 \cos \theta + a^2 = a^2 \left(\frac{1}{2} + 1\right) = \frac{3}{2} a^2.$$ Also, $$\|\vec{B}\| \|\vec{C}\| = a \times \sqrt{3} a = \sqrt{3} a^2.$$ Therefore, $$\cos \phi = \frac{\vec{B} \cdot \vec{C}}{\|\vec{B}\| \|\vec{C}\|} = \frac{\frac{3}{2} a^2}{\sqrt{3} a^2} = \frac{3}{2 \sqrt{3}} = \frac{\sqrt{3}}{2}.$$ 7. **Final angle:** $$\phi = \cos^{-1} \left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}.$$ **Answer:** The angle between $\vec{B}$ and $\vec{C}$ is $\boxed{\frac{\pi}{6}}$.