If A (-2, 1),B (2, 3),and C (-2, -4) are thre | vedclass

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(B) Let $m_1$ and $m_2$ be the slopes of $BA$ and $BC$ respectively.$m_1 = \frac{3 - 1}{2 - (-2)} = \frac{2}{4} = \frac{1}{2}$$m_2 = \frac{-4 - 3}{-2

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$	an^{-1}\left(\frac{2}{3}ight)$

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(B) Let $m_1$ and $m_2$ be the slopes of $BA$ and $BC$ respectively.$m_1 = \frac{3 - 1}{2 - (-2)} = \frac{2}{4} = \frac{1}{2}$$m_2 = \frac{-4 - 3}{-2 - 2} = \frac{-7}{-4} = \frac{7}{4}$Let $	heta$ be the angle between $BA$ and $BC$. Then,$	an 	heta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} ight| = \left| \frac{\frac{7}{4} - \frac{1}{2}}{1 + (\frac{7}{4} 	imes \frac{1}{2})} ight|$$	an 	heta = \left| \frac{\frac{5}{4}}{\frac{15}{8}} ight| = \left| \frac{5}{4} 	imes \frac{8}{15} ight| = \frac{2}{3}$Therefore,$	heta = 	an^{-1}\left(\frac{2}{3}ight)$.

If $A (-2, 1)$,$B (2, 3)$,and $C (-2, -4)$ are three points,then what is the angle between $BA$ and $BC$?

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