Question

In a $∆ABC,2a^2+4b^2+c^2=4ab+2ac$, then $\cos B$ is

• A. $0$
• B. $\frac{1}{8}$
• C. $\frac{3}{8}$
• D. $\frac{7}{8}$

Answer 1

The correct option is D

$\frac{7}{8}$

Explanation for correct option:

Step 1: Simplifying the given equation:

Given that,

$2a^2+4b^2+c^2=4ab+2ac$

$$\begin{gathered} 2a^2+4b^2+c^2=4ab+2ac \\ \Rightarrow (a^2-4ab+4b^2)+(a^2-2ac+c^2)=0 \\ \Rightarrow {(a-2b)}^2+{(a-c)}^2=0 \end{gathered}$$

Then $a-2b=0$ & $a-c=0$

Therefore, $a=2b,a=c$

Step 2: Find the value of $\cos B$:

We know that

$\begin{array}{rcl}\cos B& =& \frac{a^2+c^2-b^2}{2ac}\\ & =& \frac{a^2+a^2-{\displaystyle \frac{a^2}{4}}}{2a^2}\\ & =& \frac{7}{8}\end{array}$

Hence, the correct option is D.