Question

The sides of $\Delta ABC$ satisfy the equation $2a^2+4b^2+c^2=4ab+2ac.$ Then

• A. the triangle is isosceles
• B. the triangle is obtuse
• C. $B={\cos }^{-1}\left(7/8\right)$
• D. $A={\cos }^{-1}\left(1/4\right)$

Answer 1

Answer: (A), (C), (D)

the triangle is isosceles
$B={\cos }^{-1}\left(7/8\right)$
$A={\cos }^{-1}\left(1/4\right)$

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

$\Rightarrow \frac{a}{1}=\frac{b}{1/2}=\frac{c}{1}$

Thus, the triangle is isosceles triangle

Now, by cosine rule

$\cos B=\frac{c^2+a^2-b^2}{2ac}=\frac{1+1-\frac{1}{4}}{2}=\frac{7}{8}$

$\cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{\frac{1}{4}+1-1}{1}=\frac{1}{4}$

$\cos C=\frac{b^2+a^2-c^2}{2ab}=\frac{\frac{1}{4}+1-1}{1}=\frac{1}{4}$