The angle beta between vector overrightarrow{ | vedclass

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Question

(C) From the given figure,the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ is $60^{\circ}$.To find the angle $\beta$ between $\overrigh

Vedclass · Accepted Answer

$	an^{-1}\left(\frac{B\sin 120^{\circ}}{A+B\cos 120^{\circ}}ight)$

Vedclass · Answer

(C) From the given figure,the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ is $60^{\circ}$.To find the angle $\beta$ between $\overrightarrow{A}$ and $(\overrightarrow{A}-\overrightarrow{B})$,we consider the vector subtraction $\overrightarrow{R} = \overrightarrow{A} + (-\overrightarrow{B})$.The angle between $\overrightarrow{A}$ and $(-\overrightarrow{B})$ is $180^{\circ} - 60^{\circ} = 120^{\circ}$.Using the formula for the angle $\beta$ of the resultant vector with $\overrightarrow{A}$:$	an \beta = \frac{|-\overrightarrow{B}| \sin(120^{\circ})}{A + |-\overrightarrow{B}| \cos(120^{\circ})}$Since $|-\overrightarrow{B}| = B$,we have:$	an \beta = \frac{B \sin(120^{\circ})}{A + B \cos(120^{\circ})}$Substituting $\sin(120^{\circ}) = \frac{\sqrt{3}}{2}$ and $\cos(120^{\circ}) = -\frac{1}{2}$:$	an \beta = \frac{B(\sqrt{3}/2)}{A + B(-1/2)} = \frac{\sqrt{3}B}{2A - B}$Thus,$\beta = 	an^{-1}\left(\frac{\sqrt{3}B}{2A - B}ight)$.

The angle $\beta$ between vector $\overrightarrow{A}$ and the resultant vector $(\overrightarrow{A}-\overrightarrow{B})$ is given by:

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