A) \[\frac{{{Q}^{2}}+{{R}^{2}}}{R}\]
B) \[\frac{{{Q}^{2}}-{{R}^{2}}}{Q}\]
C) \[\frac{{{Q}^{2}}-{{R}^{2}}}{R}\]
D) \[\frac{{{Q}^{2}}+{{R}^{2}}}{Q}\]
Correct Answer: B
Solution :
Let \[\alpha \] be the angle between P and R. Then angle between P and Q is \[2\alpha \] By Lamis theorem, we have \[\frac{P}{\sin (2\pi -3\alpha )}=\frac{Q}{\sin \alpha }=\frac{R}{\sin 2\alpha }\] \[\Rightarrow \] \[\frac{P}{-\sin 3\alpha }=\frac{Q}{\sin \alpha }=\frac{R}{\sin 2\alpha }\] \[\Rightarrow \] \[\frac{P}{-(3-4si{{n}^{2}}\alpha )}=\frac{Q}{1}=\frac{R}{2\cos \alpha }\] \[\Rightarrow \]\[P=-Q(3-4{{\sin }^{2}}\alpha )\]and \[\Rightarrow \] \[\cos \alpha =\frac{R}{2Q}\] \[\Rightarrow \]\[P=-Q(4co{{s}^{2}}\alpha -1)\]and \[\cos \alpha =\frac{R}{2Q}\] \[\Rightarrow \] \[P=-Q\left( \frac{{{R}^{2}}}{{{Q}^{2}}}-1 \right)\] \[\Rightarrow \] \[P=\frac{{{Q}^{2}}-{{R}^{2}}}{Q}\]You need to login to perform this action.
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