A) \[(1,\,\sqrt{3})\]
B) \[(1,\,\sqrt{2})\]
C) \[\left( 1,\frac{\sqrt{5}+1}{2} \right)\]
D) None of these
Correct Answer: B
Solution :
Let the sides of a triangle in GP be \[\frac{a}{r},a,ar\].If \[\alpha \]is the smallest angle, then greatest angle is \[2\alpha \]. So, the side opposite to angle \[\alpha \] is \[\frac{a}{r}\] and that opposite to angle \[2\alpha \] is ar. By sine rule, \[\frac{\frac{a}{r}}{\sin \,\alpha }=\frac{ar}{\sin \,2\alpha }\] \[\Rightarrow \] \[\frac{\sin \,2\alpha }{\sin \,\alpha }={{r}^{2}}\] \[\Rightarrow \] \[2\,\cos \,\alpha ={{r}^{2}}\] Since, \[\alpha \ne 0\] \[\therefore \] \[\cos \,\alpha <1\,\,\,\,\,\Rightarrow \,\,\,\,\,2\,\cos \,\alpha <2\] \[\Rightarrow \] \[{{r}^{2}}<2\,\,\,\,\,\,\Rightarrow -\sqrt{2}<r<\sqrt{2}\] \[\Rightarrow \] \[{{r}^{2}}<2\,\,\,\,\,\,\Rightarrow -\sqrt{2}<r<\sqrt{2}\] \[(\because \,\,r>1\,given)\]
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