JEE Main & Advanced Mathematics Triangles & Properties of Triangle Question Bank Self Evaluation Test - Properties of Triangles and Height & Dstances

In a triangle \[ABC,\,\,2{{a}^{2}}+4{{b}^{2}}+{{c}^{2}}=4ab+2ac,\] then\[\cos B\] is equal to

A) 0

B) \[\frac{1}{8}\]

C) \[\frac{3}{8}\]

D) \[\frac{7}{8}\]

Correct Answer: D

Solution :

[d] \[{{a}^{2}}+{{a}^{2}}+4{{b}^{2}}-4ab=2ac-{{c}^{2}}\]

\[\Rightarrow {{(a-2b)}^{2}}+{{(a-c)}^{2}}=0\]

which is possible only when: \[a-2b=0\,\,and\,\,a-c\]

\[=0\]

or \[\frac{a}{1}=\frac{b}{1/2}=\frac{c}{1}=\lambda \] (say)

\[\therefore a=\lambda ,b=\lambda /2,c=\lambda \therefore \cos B=\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}\]

\[=\frac{{{\lambda }^{2}}+{{\lambda }^{2}}-\frac{{{\lambda }^{2}}}{4}}{2{{\lambda }^{2}}}=1-\frac{1}{8}=\frac{7}{8}\]