A) \[2\sqrt{3},\,\,4\]
B) \[\sqrt{\frac{3}{2}},\,\,2\]
C) \[3,\,\,4\]
D) \[4,\,\,5\]
Correct Answer: A
Solution :
We have,\[R=2,\,\,\theta ={{90}^{o}}\] \[\therefore \] \[{{2}^{2}}={{p}^{2}}+{{Q}^{2}}+2PQ\cos \alpha \] \[\Rightarrow \] \[4={{p}^{2}}+{{Q}^{2}}+2P(-P)\] \[\Rightarrow \] \[4={{Q}^{2}}-{{P}^{2}}\] ? (i) Also, \[P+Q\cos {{150}^{o}}=0\] \[\Rightarrow \] \[P=\frac{Q\sqrt{3}}{2}\] From Eq. (i) \[4={{Q}^{2}}-{{\left( -\frac{Q\sqrt{3}}{2} \right)}^{2}}\] \[4={{Q}^{2}}-\frac{3{{Q}^{2}}}{4}\] \[\Rightarrow \] \[4=\frac{{{Q}^{2}}}{4}\Rightarrow Q=4\] and \[P=2\sqrt{3}\]You need to login to perform this action.
You will be redirected in
3 sec