A) \[{{\cos }^{-1}}(P/Q)\]
B) \[{{\cos }^{-1}}(-P/Q)\]
C) \[{{\sin }^{-1}}\,(P/Q)\]
D) \[{{\sin }^{-1}}\,(-P/Q)\]
Correct Answer: B
Solution :
Þ \[\tan 90{}^\circ =\frac{Q\sin \theta }{P+Q\cos \theta }\]Þ \[P+Q\cos \theta =0\] \[\cos \theta =\frac{-P}{Q}\]\ \[\theta ={{\cos }^{-1}}\left( \frac{-P}{Q} \right)\]You need to login to perform this action.
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