The perimeter of $\triangle ABC$ is $6$ times the arithmetic mean of the sines of its angles. If $a = 1$, then find $A$.
By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$
$\implies$ $\sin A = \dfrac{a}{2 R}$, $\;$ $\sin B = \dfrac{b}{2 R}$, $\;$ $\sin C = \dfrac{c}{2 R}$
Perimeter of $\triangle ABC$ $= a + b + c$
Arithmetic mean of sine of angles of $\triangle ABC$ $= \dfrac{\sin A + \sin B + \sin C}{3}$
As per question,
$a + b + c = 6 \left(\dfrac{\sin A + \sin B + \sin C}{3}\right)$
i.e. $\;$ $a + b + c = 2 \left(\sin A + \sin B + \sin C\right)$
i.e. $\;$ $a + b + c = 2 \left(\dfrac{a}{2 R} + \dfrac{b}{2 R} + \dfrac{c}{2 R}\right)$ $\;\;$ [by sine rule]
i.e. $\;$ $a + b + c = \dfrac{1}{R} \left(a + b + c\right)$
$\implies$ $R = 1$ $\;$ $\because \;$ $a + b + c \neq 0$
Substituting $R = 1$, we have by sine rule,
$\dfrac{a}{\sin A} = 2 \times 1$
i.e. $\;$ $\dfrac{1}{\sin A} = 2$ $\;\;$ [Given: $a = 1$]
i.e. $\;$ $\sin A = \dfrac{1}{2}$
i.e. $\;$ $A = \sin^{-1} \left(\dfrac{1}{2}\right) = \dfrac{\pi}{6}$