1. **Problem Statement:** Prove that for any triangle ABC with usual notations, the vector magnitude relation holds: $$|\vec{a}| = |\vec{b}| \cos \gamma + |\vec{c}| \cos \beta$$ 2. **Given:** Vectors satisfy the relation $$\vec{a} + \vec{b} + \vec{c} = 0$$ which implies $$\vec{a} = -\vec{b} - \vec{c}$$ 3. **Step 1: Take dot product of both sides with $\vec{a}$:** $$\vec{a} \cdot \vec{a} = \vec{a} \cdot (-\vec{b} - \vec{c})$$ 4. **Step 2: Expand dot products:** $$|\vec{a}|^2 = -\vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{c}$$ 5. **Step 3: Express dot products in terms of magnitudes and angles:** Recall that $$\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos \theta$$ where $\theta$ is the angle between vectors $\vec{u}$ and $\vec{v}$. Here, angles between $\vec{a}$ and $\vec{b}$ is $180^\circ - \gamma$, and between $\vec{a}$ and $\vec{c}$ is $180^\circ - \beta$. So, $$|\vec{a}|^2 = -|\vec{a}| |\vec{b}| \cos(180^\circ - \gamma) - |\vec{a}| |\vec{c}| \cos(180^\circ - \beta)$$ 6. **Step 4: Use the identity $\cos(180^\circ - \theta) = -\cos \theta$:** $$|\vec{a}|^2 = -|\vec{a}| |\vec{b}| (-\cos \gamma) - |\vec{a}| |\vec{c}| (-\cos \beta)$$ which simplifies to $$|\vec{a}|^2 = |\vec{a}| |\vec{b}| \cos \gamma + |\vec{a}| |\vec{c}| \cos \beta$$ 7. **Step 5: Divide both sides by $|\vec{a}|$ (assuming $|\vec{a}| \neq 0$):** $$|\vec{a}| = |\vec{b}| \cos \gamma + |\vec{c}| \cos \beta$$ **Final answer:** $$\boxed{|\vec{a}| = |\vec{b}| \cos \gamma + |\vec{c}| \cos \beta}$$ This completes the proof using vector dot product properties and angle identities.