(c)
$\theta = \tan^{- 1}\left\{ \frac{\frac{2}{3} \times \frac{4}{3} - 1}{\frac{2}{3} + \frac{4}{3}} \right\}$
$\Rightarrow \theta = \tan^{- 1}{\left\{ - \frac{\frac{1}{9}}{2} \right\} = \tan^{- 1}{\left\{ - \frac{1}{18} \right\}\ }}$
$\therefore Angle\ is\ acute\ angle.$
$\therefore\$k=$\frac{1}{18}$