Question

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

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

Answer 1

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

The correct options are
A The triangle is isosceles
C $B={\cos }^{-1}\left(\frac{7}{8}\right)$
D $A={\cos }^{-1}\left(\frac{1}{4}\right)$

$(a^2-2ac+c^2)+(a^2-4ab+4b^2)=0$
Or, $(a-c{)}^2+(a-2b{)}^2=0$
$\Rightarrow a=c$ and $a=2b$
Therefore, the triangle is isosceles.
Also, $\cos B=\frac{a^2+c^2-b^2}{2ac}=\frac{7b^2}{8b^2}=\frac{7}{8}$
$\cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{1}{4}$