Question

In a triangle $ABC,2a^2+4b^2+c^2=4ab+2ac$ 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 $\frac{7}{8}$

In a triangle $ABC,2a^2+4b^2+c^2=4ab+2ac$
$\Rightarrow a^2+a^2+{\left(2b\right)}^2+c^2-4ab-2ac=0$
$\Rightarrow {(a-c)}^2+{(2b-a)}^2=0$ which is possible when $a-2b=0$ and $a-c=0$
or $a=2b$ and $a=c$
or $\frac{a}{1}=\frac{b}{\frac{1}{2}}=\frac{c}{1}=\lambda$(say)
$\therefore a=\lambda ;b=\frac{\lambda }{2};c=\lambda$
Substituting the values of $a,b$ and $c$ in cosine angle formulae
$\cos B=\frac{c^2+a^2-b^2}{2ac}$ we get
$\cos B=\frac{{\lambda }^2+{\lambda }^2-{\left(\frac{\lambda }{2}\right)}^2}{2ac}$
$=\frac{{\lambda }^2(1+1-\frac{1}{4})}{2{\lambda }^2}$
$=\frac{1}{2}\times \frac{7}{4}=\frac{7}{8}$