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The perimeter of a $\triangle A B C$ is 6 times the arithmetic mean of the sine of its angles. If its side $B C$ is of unit length, then $\angle A=$
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Verified Answer
The correct answer is:
$\frac{\pi}{6}$
It is given that length of side $B C=1$ unit, then according to the given information
$a+b+c=6 \frac{\sin A+\sin B+\sin C}{3}$
$\Rightarrow a+b+c=2(\sin A+\sin B+\sin C)$
$\Rightarrow \quad 2 R(\sin A+\sin B+\sin C)=2$ $(\sin A+\sin B+\sin C) \quad($ from sine law $)$
$\Rightarrow \quad 2 R=2$
Now, from sine law, we have
$\frac{a}{\sin A}=2 R \Rightarrow \frac{1}{\sin A}=2 \Rightarrow \sin A=\frac{1}{2} \Rightarrow A=\frac{\pi}{6}$
$a+b+c=6 \frac{\sin A+\sin B+\sin C}{3}$
$\Rightarrow a+b+c=2(\sin A+\sin B+\sin C)$
$\Rightarrow \quad 2 R(\sin A+\sin B+\sin C)=2$ $(\sin A+\sin B+\sin C) \quad($ from sine law $)$
$\Rightarrow \quad 2 R=2$
Now, from sine law, we have
$\frac{a}{\sin A}=2 R \Rightarrow \frac{1}{\sin A}=2 \Rightarrow \sin A=\frac{1}{2} \Rightarrow A=\frac{\pi}{6}$
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