A) \[\frac{1}{9}\]
B) \[\frac{2}{9}\]
C) \[\frac{1}{18}\]
D) \[\frac{1}{3}\]
E) \[\frac{8}{9}\]
Correct Answer: C
Solution :
We know, \[\theta ={{\tan }^{-1}}\left( \frac{{{b}_{xy}}\times {{b}_{xy}}-1}{{{b}_{yx}}+{{b}_{xy}}} \right)\] \[\theta ={{\tan }^{-1}}\left\{ \frac{\frac{2}{3}\times \frac{4}{3}-1}{\frac{2}{3}+\frac{4}{3}} \right\}\] \[\Rightarrow \] \[={{\tan }^{-1}}\left\{ -\frac{1/9}{2} \right\}\] \[={{\tan }^{-1}}\left\{ -\frac{1}{18} \right\}\] \[\because \]Angle is acute angle \[\therefore \] \[k=\frac{1}{18}\]
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