With usual notations,the perimeter of a trian | vedclass

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Question

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

Vedclass · Accepted Answer

$\frac{\pi}{6}$

Vedclass · Answer

(A) The perimeter of the triangle is $P = a + b + c$. 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,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$,where $R$ is the circumradius.Substituting these into the equation: $2R(\sin A + \sin B + \sin C) = 2(\sin A + \sin B + \sin C)$.This implies $2R = 2$,so $R = 1$.Since $a = 2R \sin A$ and $a = 1$,we have $1 = 2(1) \sin A$.Therefore,$\sin A = \frac{1}{2}$,which gives $A = \frac{\pi}{6}$ or $A = \frac{5\pi}{6}$.Given the standard context of such problems,we select $A = \frac{\pi}{6}$.

With usual notations,the perimeter of a triangle $ABC$ is $6$ times the arithmetic mean of the sines of its angles. If $a=1$,then $\angle A=$

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