If A→=B→−C→ then determine the angle between A→ and B→

θ = cos − 1 [ A 2 + B 2 − C 2 2 AB ]

θ = cos − 1 [ A 2 + B 2 − C 2 2 AB ]

θ = sin − 1 [ A 2 + B 2 − C 2 2 AB ]

θ = tan − 1 [ A 2 + B 2 − C 2 2 AB ]

θ = cot − 1 [ A 2 + B 2 − C 2 2 AB ]

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If $\vec{A} = \vec{B} - \vec{C}$ then determine the angle between $\vec{A} and \vec{B}$

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θ = \cos^{- 1} [\frac{A^{2} + B^{2} - C^{2}}{2 AB}]

θ = \sin^{- 1} [\frac{A^{2} + B^{2} - C^{2}}{2 AB}]

θ = \tan^{- 1} [\frac{A^{2} + B^{2} - C^{2}}{2 AB}]

θ = \cot^{- 1} [\frac{A^{2} + B^{2} - C^{2}}{2 AB}]

answer is A.

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\bar{A} = \bar{B} - \bar{C}

$\bar{C} = \bar{B} - \bar{A}$

Let $θ$ be angle between $\bar{A} and \bar{B}$

$C^{2} = A^{2} + B^{2} + 2 AB \cos (180 - θ)$

$θ = \cos^{- 1} [\frac{A^{2} + B^{2} - C^{2}}{2 AB}]$

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