In a triangle ABC,sin A : sin B : sin C = 1 : | vedclass

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(C) By the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$. Thus,$a : b : c = \sin A : \sin B : \sin C = 1 : 2 : 3$. Let $a = x

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$12$

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(C) By the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$. Thus,$a : b : c = \sin A : \sin B : \sin C = 1 : 2 : 3$. Let $a = x$,$b = 2x$,and $c = 3x$. Given $b = 4 \, 	ext{cm}$,we have $2x = 4$,which implies $x = 2 \, 	ext{cm}$. Therefore,$a = 2 \, 	ext{cm}$,$b = 4 \, 	ext{cm}$,and $c = 6 \, 	ext{cm}$. The perimeter of the triangle is $a + b + c = 2 + 4 + 6 = 12 \, 	ext{cm}$.

In a triangle $ABC$,$\sin A : \sin B : \sin C = 1 : 2 : 3$. If $b = 4 \, \text{cm}$,the perimeter of the triangle is ..... $\text{cm}$.

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