A) \[\sqrt{\frac{9{{a}^{2}}-{{c}^{2}}}{8{{a}^{2}}}}\]
B) \[\sqrt{\frac{9{{a}^{2}}-{{c}^{2}}}{8{{c}^{2}}}}\]
C) \[\sqrt{\frac{9{{a}^{2}}+{{c}^{2}}}{8{{a}^{2}}}}\]
D) None of these
Correct Answer: B
Solution :
We have \[\cos A=\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}\] \[\Rightarrow \] \[{{b}^{2}}-2bc\cos A+({{c}^{2}}-{{a}^{2}})=0\] It is given that \[{{b}_{1}}\] and \[{{b}_{2}}\] are roots of this equation. Therefore \[{{b}_{1}}+{{b}_{2}}\] = \[2c\cos A\] and \[{{b}_{1}}{{b}_{2}}={{c}^{2}}-{{a}^{2}}\] \[\Rightarrow \]\[3{{b}_{1}}=2c\cos A\], \[2b_{1}^{2}={{c}^{2}}-{{a}^{2}}\], \[(\because \,\,{{b}_{2}}=2{{b}_{1}}\] given) \[\Rightarrow \] \[2\,{{\left( \frac{2c}{3}\cos A \right)}^{2}}={{c}^{2}}-{{a}^{2}}\Rightarrow 8{{c}^{2}}(1-{{\sin }^{2}}A)=9{{c}^{2}}-9{{a}^{2}}\] \[\Rightarrow \] \[\sin A=\sqrt{\frac{9{{a}^{2}}-{{c}^{2}}}{8{{c}^{2}}}}\].
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