A) \[{{\cos }^{-1}}\left( \frac{P}{Q} \right)\]
B) \[\pi -{{\cos }^{-1}}\left( \frac{Q}{P} \right)\]
C) \[\pi -{{\cos }^{-1}}\left( \frac{P}{Q} \right)\]
D) None of these
Correct Answer: B
Solution :
\[\tan \,{{90}^{o}}=\frac{P\,\,\sin \,\alpha }{Q+P\,\,\cos \,\,\alpha }\] \[\Rightarrow \] \[Q+P\,\,\cos \,\alpha =0\] \[\Rightarrow \] \[\cos \,\alpha =-\frac{Q}{P}\] \[\Rightarrow \] \[\alpha ={{\cos }^{-1}}\left( -\frac{Q}{P} \right)\] \[\Rightarrow \] \[\alpha =\pi -{{\cos }^{-1}}\left( \frac{Q}{P} \right)\]You need to login to perform this action.
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