A) \[2b\]
B) \[2c\]
C) \[3a\]
D) \[x+y-1=0\]
Correct Answer: D
Solution :
If \[\text{A}:\text{B}:\text{C}=\text{3}:\text{5}:\text{4}\] i.e.\[A=3x,\text{ }B=5x,\text{ }C=4x\]. Then, \[\text{A}+\text{B}+\text{C}=\text{18}0{}^\circ \] \[\Rightarrow \]\[3x+5x+4x={{180}^{o}}\] \[\Rightarrow \] \[12x={{180}^{o}}\] \[\Rightarrow \] \[x={{15}^{o}}\] \[\therefore \]\[A=3\times {{15}^{o}}={{45}^{o}},B=5\times {{15}^{^{o}}}={{75}^{o}}\] and \[C=4\times {{15}^{o}}={{60}^{o}}\] Now\[\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{C}{\sin C}\]=2R \[\Rightarrow \] \[a=2R\sin {{45}^{o}}=\frac{2R}{\sqrt{2}}=\sqrt{2}R\] \[b=2R\sin {{75}^{o}}=2R\left( \frac{\sqrt{3}+1}{2\sqrt{2}} \right)=\frac{\sqrt{3}+1}{\sqrt{2}}R\] And \[c=2R\sin {{60}^{o}}=2R\left( \frac{\sqrt{3}}{2} \right)=\sqrt{3}R\] \[\therefore \]\[a+b+\sqrt{2}c=\sqrt{2}R+\frac{\sqrt{3}+1}{\sqrt{2}}R+\sqrt{2}\times \sqrt{3}R\] \[=\frac{2R+\sqrt{3}R+R+2\sqrt{3}R}{\sqrt{2}}=\frac{3R+3\sqrt{3}R}{\sqrt{2}}\] \[=\frac{3R(\sqrt{3+1})}{\sqrt{2}}=3b\]
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