If vec{A} + vec{B} = vec{C} and |vec{A}| = |v | vedclass

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(D) Given that $\vec{A} + \vec{B} = \vec{C}$ and $|\vec{A}| = |\vec{B}| = |\vec{C}| = A$ (let).Squaring both sides of the vector equation $\vec{A} + \

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$120$

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(D) Given that $\vec{A} + \vec{B} = \vec{C}$ and $|\vec{A}| = |\vec{B}| = |\vec{C}| = A$ (let).Squaring both sides of the vector equation $\vec{A} + \vec{B} = \vec{C}$,we get:$|\vec{A} + \vec{B}|^2 = |\vec{C}|^2$$|\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}| \cos 	heta = |\vec{C}|^2$,where $	heta$ is the angle between $\vec{A}$ and $\vec{B}$.Substituting the magnitudes: $A^2 + A^2 + 2A^2 \cos 	heta = A^2$.$2A^2 + 2A^2 \cos 	heta = A^2$.$2A^2 \cos 	heta = A^2 - 2A^2 = -A^2$.$\cos 	heta = -A^2 / 2A^2 = -1/2$.Since $\cos 	heta = -1/2$,the angle $	heta = 120^o$.

If $\vec{A} + \vec{B} = \vec{C}$ and $|\vec{A}| = |\vec{B}| = |\vec{C}|$,then the angle between $\vec{A}$ and $\vec{B}$ is ....... $^o$.

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