Question

The perimeter of a $\Delta ABC$ is $6$ times the arithmetic mean of the sines of its angle. If the side $b$ is $1$, then evaluate $\underset{A\to C}{lim}\frac{\sqrt{1-4(2-\sqrt{3})\sin A\sin C}}{|A-C|}$

Answer 1

Given $a+b+c=6\left(\frac{\sin A+\sin B+\sin C}{3}\right)\Rightarrow a+b+c=2(\sin A+\sin B+\sin C)$ ...(1)

$\because \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\Rightarrow a=\frac{\sin A}{\sin B};c=\frac{\sin C}{\sin B}(\because b=1)$

$\therefore$ From (1) $\frac{\sin A}{\sin B}+1+\frac{\sin C}{\sin B}=2(\sin A+\sin B+\sin C)$

$\Rightarrow (\sin A+\sin B+\sin C)=2(\sin A+\sin B+\sin C)\sin B\Rightarrow (\sin A+\sin B+\sin C)(2\sin B-1)=0$

$\Rightarrow 2\sin B-1=0$ $[\because$ In $△,\sin A,\sin B,\sin C>0,\sin A+\sin B+\sin C\ne 0]$

$\Rightarrow \sin B=\frac{1}{2}\Rightarrow \cos B=\frac{\sqrt{3}}{2}$

$\therefore$ By cosine formula

$\cos B=\frac{a^2+c^2-b^2}{2ac}\Rightarrow \frac{\sqrt{3}}{2}=\frac{a^2+c^2-b^2}{2ac}\Rightarrow \sqrt{3}=\frac{a^2+c^2-b^2}{ac}\Rightarrow \sqrt{3}ac=a^2+c^2-b^2\Rightarrow \sqrt{3}ac-2ac=a^2+c^2-2ac-b^2\Rightarrow (\sqrt{3}-2)ac={(a-c)}^2-b^2\Rightarrow {(a-c)}^2=b^2-(2-\sqrt{3})ac\Rightarrow |a-c|=\sqrt{b^2-(2-\sqrt{3})ac}\Rightarrow |2R\sin A-2R\sin C|=\sqrt{4R^2{\sin }^{2}B-(2-\sqrt{3})4R^2\sin A\sin C}\Rightarrow |\sin A-\sin C|=\sqrt{{\sin }^{2}B-(2-\sqrt{3})\sin A\sin C}$

$\Rightarrow \left|2\cos \frac{A+C}{2}\sin \frac{A-C}{2}\right|=\sqrt{\frac{1}{4}-(2-\sqrt{3})\sin A\sin C}$

$\Rightarrow 2\cos \frac{A+C}{2}\left|\sin \frac{A-C}{2}\right|=\frac{1}{2}\sqrt{1-(2-\sqrt{2})\sin A\sin C}$

$\Rightarrow \sqrt{1-4(2-\sqrt{3})\sin A\sin C}=4\cos \frac{A+C}{2}\left|\sin \frac{A-C}{2}\right|$

$\Rightarrow \sqrt{\frac{1-4(2-\sqrt{3})\sin A\sin C}{|A-C|}}=\frac{4\cos \frac{A+C}{2}\left|\sin \frac{A-C}{2}\right|}{|A-C|}$

$\Rightarrow \underset{A\to C}{lim}\sqrt{\frac{1-4(2-\sqrt{3})\sin A\sin C}{|A-C|}}=\underset{A\to C}{lim}\frac{4\cos \frac{A+C}{2}\left|\sin \frac{A-C}{2}\right|}{|A-C|}=2$