Given that vec{A} + vec{B} = vec{C}. If |vec{ | vedclass

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(B) The magnitude of the resultant vector $\vec{C} = \vec{A} + \vec{B}$ is given by the formula: $|\vec{C}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}|

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$60$

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(B) The magnitude of the resultant vector $\vec{C} = \vec{A} + \vec{B}$ is given by the formula: $|\vec{C}|^2 = |\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}| \cos 	heta$,where $	heta$ is the angle between $\vec{A}$ and $\vec{B}$.Given values are $|\vec{A}| = 4$,$|\vec{B}| = 5$,and $|\vec{C}| = \sqrt{61}$.Substituting these values into the equation:$(\sqrt{61})^2 = 4^2 + 5^2 + 2(4)(5) \cos 	heta$$61 = 16 + 25 + 40 \cos 	heta$$61 = 41 + 40 \cos 	heta$$61 - 41 = 40 \cos 	heta$$20 = 40 \cos 	heta$$\cos 	heta = \frac{20}{40} = \frac{1}{2}$Since $\cos 	heta = \frac{1}{2}$,the angle $	heta = 60^o$.

Given that $\vec{A} + \vec{B} = \vec{C}$. If $|\vec{A}| = 4, |\vec{B}| = 5$ and $|\vec{C}| = \sqrt{61}$,the angle between $\vec{A}$ and $\vec{B}$ is ....... $^o$

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