A) 0
B) \[\frac{1}{8}\]
C) \[\frac{3}{8}\]
D) \[\frac{7}{8}\]
Correct Answer: D
Solution :
\[2{{a}^{2}}+4{{b}^{2}}+{{c}^{2}}=4ab+2ac\] \[\Rightarrow \]\[{{a}^{2}}+{{(2b)}^{2}}-4ab+{{a}^{2}}+{{c}^{2}}-2ac=0\] \[\Rightarrow \]\[{{(a-2b)}^{2}}+{{(a-c)}^{2}}=0\] \[\Rightarrow \]\[a=2b=c\] \[\cos B=\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}\] \[=\frac{{{c}^{2}}+{{c}^{2}}-{{\left( \frac{c}{2} \right)}^{2}}}{2\times c\times c}=\frac{2{{c}^{2}}-\frac{{{c}^{2}}}{4}}{2{{c}^{2}}}\] \[\Rightarrow \] \[\cos B=\frac{7}{8}\]
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