Question

The sides of a $△ABC$ satisfy the equation, $2a^2+4b^2+c^2=4ab+2ac,$ then find $\angle A$ and $\angle B$

• A. $A={\cos }^{-1}\left(\frac{1}{4}\right)$
• B. $B={\cos }^{-1}\left(\frac{7}{8}\right)$
• C. $A={\cos }^{-1}\left(\frac{1}{3}\right)$
• D. $B={\cos }^{-1}\left(\frac{1}{8}\right)$

Answer 1

Answer: (A), (B)

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

Given expression ${(a-c)}^2+{(a-2b)}^2=0$

$\Rightarrow a=2b$ and $c=a$.

Sides are $2b,b,2b$

$△$ is isosceles and $\cos B=\frac{7}{8}$ and $\cos A=\frac{1}{4}$

$\because \cos B=\frac{a^2+c^2-b^2}{2ac}=\frac{7b^2}{8b^2}=\frac{7}{8}\Rightarrow B={\cos }^{-1}\left(\frac{7}{8}\right)$

and $\cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{1b^2}{4b^2}=\frac{1}{4}\Rightarrow A={\cos }^{-1}\left(\frac{1}{4}\right)$