With usual notations, perimeter of a triangle $\mathrm{ABC}$ is 6 times the arithmetic mean of since of its…
With usual notations, perimeter of a triangle $\mathrm{ABC}$ is 6 times the arithmetic mean of since of its angles. If $a=1$, then measure of angle $\mathrm{A}=$
$\frac{\pi^c}{3}$
$\frac{\pi^c}{2}$
$\frac{\pi^c}{4}$
$\frac{\pi^c}{6}$
Solution
$\begin{aligned}
& \text { Let } \frac{\mathrm{a}}{\sin \mathrm{A}}=\frac{\mathrm{b}}{\sin \mathrm{B}}=\frac{\mathrm{c}}{\sin \mathrm{C}}=\mathrm{k} \\
& \therefore \sin \mathrm{A}=\frac{\mathrm{a}}{\mathrm{k}}, \sin \mathrm{B}=\frac{\mathrm{b}}{\mathrm{k}}, \sin \mathrm{C}=\frac{\mathrm{c}}{\mathrm{k}}
\end{aligned}$
With usual notations, from the given data, we write
$\begin{aligned}
& \mathrm{a}+\mathrm{b}+\mathrm{c}=6\left[\frac{\left(\frac{\mathrm{a}}{\mathrm{k}}+\frac{\mathrm{b}}{\mathrm{k}}+\frac{\mathrm{c}}{\mathrm{k}}\right)}{3}\right] \\
& \therefore(\mathrm{a}+\mathrm{b}+\mathrm{c})=\frac{2(\mathrm{a}+\mathrm{b}+\mathrm{c})}{\mathrm{k}} \Rightarrow \mathrm{k}=2 \\
& \therefore \sin \mathrm{A}=\frac{\mathrm{a}}{\mathrm{k}}=\frac{1}{2} \Rightarrow \mathrm{A}=\frac{\pi^{\mathrm{c}}}{6}
\end{aligned}$