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(A) The perimeter of $\Delta ABC$ is $a + b + c$. The arithmetic mean of the sines of its angles is $\frac{\sin A + \sin B + \sin C}{3}$.Given: $a + b

Vedclass · Accepted Answer

$\frac{\pi}{6}$

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(A) The perimeter of $\Delta ABC$ is $a + b + c$. The arithmetic mean of the sines of its angles is $\frac{\sin A + \sin B + \sin C}{3}$.Given: $a + b + c = 6 	imes \frac{\sin A + \sin B + \sin C}{3} = 2(\sin A + \sin B + \sin C)$.Using the Sine Rule,$a = k \sin A$,$b = k \sin B$,and $c = k \sin C$,where $k = 2R$ (the diameter of the circumcircle).Substituting these into the equation: $k(\sin A + \sin B + \sin C) = 2(\sin A + \sin B + \sin C)$.Since $\sin A + \sin B + \sin C 
eq 0$ for a triangle,we have $k = 2$.Given $a = 1$,we use $a = k \sin A$:$1 = 2 \sin A \implies \sin A = \frac{1}{2}$.Since $A$ is an angle of a triangle,$A = \frac{\pi}{6}$ or $A = \frac{5\pi}{6}$. Given the options,$A = \frac{\pi}{6}$.

The perimeter of a $\Delta ABC$ is $6$ times the arithmetic mean of the sines of its angles. If the side $a$ is $1$,then the angle $A$ is

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