Question

In a triangle ABC, $2a^2+4b^2+6c^2=4ab+2ac$, then the numerical value of $\cos B$ is equal to

• A. $0$
• B. $\frac{3}{8}$
• C. $\frac{5}{8}$
• D. $\frac{7}{8}$

Answer 1

The correct option is D $\frac{7}{8}$

$2a^2+4b^2+6c^2=4ab+2ac$
$\Rightarrow (a-2b{)}^2+(a-c{)}^2=0$
Which is possibel only when $a-2b=0$ and $a-c=0$
or $\frac{a}{1}=\frac{b}{\left(\frac{1}{2}\right)}=\frac{c}{1}=\lambda$ (say)
$\therefore a=\lambda ,b=\frac{\lambda }{2},C=\lambda$
$\therefore \cos B=\frac{a^2=c^2-b^2}{2ac}=\frac{{\lambda }^2+{\lambda }^2-\frac{{\lambda }^2}{4}}{2{\lambda }^2}$
$=1-\frac{1}{8}=\frac{7}{8}$