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Question

(A) The perimeter of $	riangle 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

Vedclass · Accepted Answer

$\frac{\pi}{6}$

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(A) The perimeter of $	riangle ABC$ is $a+b+c$. The arithmetic mean of the sines of its angles is $\frac{\sin A + \sin B + \sin C}{3}$.Given: $a+b+c = 6 	imes \left(\frac{\sin A + \sin B + \sin C}{3}ight) = 2(\sin A + \sin B + \sin C)$.Using the Sine Rule,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$,we have:$2R(\sin A + \sin B + \sin C) = 2(\sin A + \sin B + \sin C)$.This implies $2R = 2$,so $R = 1$.Given $BC = a = 1$,we use $a = 2R \sin A$:$1 = 2(1) \sin A \implies \sin A = \frac{1}{2}$.Since $A$ is an angle of a triangle,$A = 30^{\circ} = \frac{\pi}{6}$.

The perimeter of a $\triangle ABC$ is $6$ times the arithmetic mean of the values of the sine of its angles. If its side $BC$ is of unit length,then $\angle A=$

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