In $ riangle \mathrm{ABC}$, if $\cos \mathrm{A}+\cos \mathrm{C}=4 \sin ^2 \frac{\mathrm{B}}{2}$, then the ratio between the perimeter of the triangle and $(a+c)$ is

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In $	riangle \mathrm{ABC}$, if $\cos \mathrm{A}+\cos \mathrm{C}=4 \sin ^2 \frac{\mathrm{B}}{2}$, then the ratio between the perimeter of the triangle and $(a+c)$ is

Aicharya · Accepted Answer

Given $\cos A+\cos C=4 \sin ^2 \frac{B}{2}$ $\begin{aligned} & \Rightarrow 2 \cos \left(\frac{A+C}{2}ight)+\cos \left(\frac{A-C}{2}ight)=4 \sin ^2 \frac{B}{2} \ & \Rightarrow \cos \left(\frac{A-C}{2}ight)=2 \sin \frac{B}{2} \ & \Rightarrow \cos \frac{A}{2} \cdot \cos \frac{C}{2}+\sin \frac{A}{2} \sin \frac{C}{2}=2 \sin \frac{B}{2} \ & \Rightarrow \sqrt{\frac{s(s-a)}{b c}} \sqrt{\frac{s(s-c)}{a b}}+\sqrt{\frac{(s-b)(s-c)}{b c}} \sqrt{\frac{(s-a)(s-b)}{a b}} \ & =2 \sqrt{\frac{(s-a)(s-c)}{a c}} \ & \Rightarrow \frac{s}{b}+\frac{s-b}{b}=2 \ & \Rightarrow a+c=2 b \end{aligned}$ Now, $\frac{a+b+c}{a+c}=\frac{2 b+b}{2 b}=\frac{3}{2}=3: 2$

In $\triangle \mathrm{ABC}$, if $\cos \mathrm{A}+\cos \mathrm{C}=4 \sin ^2 \frac{\mathrm{B}}{2}$, then the…

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