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(D) The slope of line $AB$ $(m_1)$ is $\frac{-2 - 2}{12 - (-4)} = \frac{-4}{16} = -\frac{1}{4}$.The slope of line $BC$ $(m_2)$ is $\frac{6 - (-2)}{8 -

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$	an^{-1}\left(\frac{7}{6}ight)$

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(D) The slope of line $AB$ $(m_1)$ is $\frac{-2 - 2}{12 - (-4)} = \frac{-4}{16} = -\frac{1}{4}$.The slope of line $BC$ $(m_2)$ is $\frac{6 - (-2)}{8 - 12} = \frac{8}{-4} = -2$.The angle $	heta$ between two lines with slopes $m_1$ and $m_2$ is given by $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$.Substituting the values: $	an 	heta = \left| \frac{-1/4 - (-2)}{1 + (-1/4)(-2)} ight| = \left| \frac{-1/4 + 2}{1 + 1/2} ight| = \left| \frac{7/4}{3/2} ight| = \frac{7}{4} 	imes \frac{2}{3} = \frac{7}{6}$.Therefore,$\angle B = 	an^{-1}\left(\frac{7}{6}ight)$.

If the coordinates of the vertices $A, B, C$ of the triangle $ABC$ are $(-4, 2)$,$(12, -2)$,and $(8, 6)$ respectively,then $\angle B$ =

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