With usual notations,the perimeter of a trian | vedclass

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(D) Let $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$. Then $\sin A = \frac{a}{k}$,$\sin B = \frac{b}{k}$,and $\sin C = \frac{c}{k}$. T

Vedclass · Accepted Answer

$\frac{\pi^c}{6}$

Vedclass · Answer

(D) Let $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$. Then $\sin A = \frac{a}{k}$,$\sin B = \frac{b}{k}$,and $\sin C = \frac{c}{k}$. The perimeter of the triangle is $a+b+c$. The arithmetic mean of the sines of its angles is $\frac{\sin A + \sin B + \sin C}{3}$. According to the problem,$a+b+c = 6 	imes \left( \frac{\sin A + \sin B + \sin C}{3} ight)$. Substituting the values,we get $a+b+c = 2(\sin A + \sin B + \sin C) = 2 \left( \frac{a+b+c}{k} ight)$. Thus,$1 = \frac{2}{k}$,which implies $k = 2$. Since $\sin A = \frac{a}{k}$ and $a=1$,we have $\sin A = \frac{1}{2}$. Therefore,$A = \frac{\pi^c}{6}$.

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 the measure of angle $A$ is:

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