A) \[{{\tan }^{-1}}\left( -\frac{P}{Q} \right)\]
B) \[{{\tan }^{-1}}\left( \frac{P}{Q} \right)\]
C) \[{{\sin }^{-1}}\left( \frac{P}{Q} \right)\]
D) \[{{\cos }^{-1}}\left( -\frac{P}{Q} \right)\]
Correct Answer: D
Solution :
Let \[\theta \] be the angle which resultant (R) makes with forces P and Q, then\[\tan \theta =\frac{Q\sin \alpha }{P+Q\cos \alpha }\]where a is angle between the forces P and Q. Given, \[\theta ={{90}^{o}}\] \[\therefore \] \[\tan {{90}^{o}}=\frac{Q\sin \alpha }{P+Q\cos \alpha }\] \[\Rightarrow \] \[\infty =\frac{Q\sin \alpha }{P+Q\cos \alpha }\] \[\Rightarrow \] \[P+Q\cos \alpha =0\] \[\Rightarrow \] \[\cos \alpha =-\frac{P}{Q}\] \[\Rightarrow \] \[\alpha =-{{\cos }^{-1}}\left( -\frac{P}{Q} \right)\]
You need to login to perform this action.
You will be redirected in
3 sec
