A) \[{{\tan }^{-1}}\left( \frac{2}{3} \right)\]
B) \[{{\tan }^{-1}}\left( \frac{3}{2} \right)\]
C) \[{{\tan }^{-1}}\left( \frac{1}{3} \right)\]
D) \[{{\tan }^{-1}}\left( \frac{1}{2} \right)\]
Correct Answer: A
Solution :
\[A\,\left( -\,2,1 \right),\] \[B\,\left( 2,\,\,3 \right)\] and \[C\left( -2,-4 \right)\]are three given points slope of the line BA \[{{m}_{1}}=\frac{1-3}{-2-2}=\frac{1}{2}\] \[\left( \text{using slope formula}\,m=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \right)\] Slope of the line BC \[{{m}_{2}}=\frac{-\,4-3}{-2-2}=\frac{7}{4}\] Now, angle between AB and BC is given by \[\tan \theta =\left| \frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|\] \[=\left| \frac{\frac{1}{2}-\frac{7}{4}}{1+\frac{1}{2}\cdot \frac{7}{4}} \right|\] \[\Rightarrow \] \[\tan \theta =\left| -\frac{10}{15} \right|\] \[\Rightarrow \] \[\tan \theta =\left| -\frac{2}{3} \right|\] \[\Rightarrow \] \[\theta ={{\tan }^{-1}}\left| -\frac{2}{3} \right|\] \[\Rightarrow \] \[\theta ={{\tan }^{-1}}\left( \frac{2}{3} \right)\] \[[\because \left| -x \right|=x]\]
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