A) \[{{\tan }^{-1}}2\]
B) \[{{\tan }^{-1}}\frac{1}{3}\]
C) \[{{\tan }^{-1}}\frac{1}{2}\]
D) \[{{\tan }^{-1}}(-2)\]
E) \[{{\tan }^{-1}}(-3)\]
Correct Answer: E
Solution :
Centroid of a triangle whose vertices are (0,0), \[(\cos \theta ,\sin \theta )\]and\[(\sin \theta ,-\cos \theta )\]is \[\left( \frac{\cos \theta +\sin \theta }{3},\frac{\sin \theta -\cos \theta }{3} \right)\] But centroid lies on a line\[y=2x\] \[\therefore \] \[\frac{-\cos \theta +\sin \theta }{3}=\frac{2(\sin \theta +\cos \theta )}{3}\] \[\Rightarrow \] \[-3\cos \theta =\sin \theta \] \[\Rightarrow \] \[\tan \theta =-3\] \[\Rightarrow \] \[\theta ={{\tan }^{-1}}(-3)\]
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