Question

In a $△ABC,2\cos \left(\frac{A-C}{2}\right)=\frac{a+c}{\sqrt{a^2+c^2-ac}},$ then

• A. $B=\frac{\pi }{3}$
• B. $B=C$
• C. $A,B,C$ are in A.P.
• D. $B+C=A$

Answer 1

The correct option is A $B=\frac{\pi }{3}$

$2\cos \left(\frac{A-C}{2}\right)=\frac{2\sin \left(\frac{A+C}{2}\right)\cos \left(\frac{A-C}{2}\right)}{\sin \left(\frac{A+C}{2}\right)}$
$=\frac{\sin A+\sin C}{\sin \left(\frac{A+C}{2}\right)}$
$\cos B=\frac{a^2+c^2-b^2}{2ac}$ if $\angle B={60}^0$
Then $\frac{a^2+c^2-b^2}{2ac}=\cos {60}^0$
$\Rightarrow \frac{a^2+c^2-b^2}{2ac}=\frac{1}{2}$
$\Rightarrow a^2+c^2-b^2=ac$
$\Rightarrow a^2+c^2-ac=b^2$
$=\frac{a+c}{k\sin {60}^0}=\frac{a+c}{b}$
$=\frac{a+c}{\sqrt{a^2+c^2-ac}}$