The angle between vector $\left( {\overrightarrow A } \right)$ and $\left( {\overrightarrow A - \overrightarrow B } \right)$ is :

Question

${\tan ^{ - 1}}\left( {{{ - {B \over 2}} \over {A - B{{\sqrt 3 } \over 2}}}} \right)$

${\tan ^{ - 1}}\left( {{A \over {0.7B}}} \right)$

${\tan ^{ - 1}}\left( {{{\sqrt 3 B} \over {2A - B}}} \right)$

${\tan ^{ - 1}}\left( {{{B\cos \theta } \over {A - B\sin \theta }}} \right)$

Please login to view the detailed solution steps...