The drameter length of the circumcircle of $$ riangle \mathrm{ABC} $$ in which a $$ =8 \mathrm{sin} \mathrm{Acm} $$ equals $$ \begin{array}{llll} ext { a) } 4 & \mathrm{~cm}\end{array} $$

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The drameter length of the circumcircle of $$ 	riangle \mathrm{ABC} $$ in which a $$ =8 \mathrm{sin} \mathrm{Acm} $$ equals $$ \begin{array}{llll}	ext { a) } 4 & \mathrm{~cm}\end{array} $$

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The diameter of the circumcircle is $$ 8 \text{ cm} $$. To find the diameter of the circumcircle of triangle $$ \triangle ABC $$, we can use the formula for the circumradius $$ R $$ of a triangle, which is given by: $$ R = \frac{a}{2 \sin A} $$ where $$ a $$ is the length of side $$ a $$ opposite angle $$ A $$. From the problem, we have: $$ a = 8 \sin A \text{ cm} $$ Substituting this into the formula for $$ R $$: $$ R = \frac{8 \sin A}{2 \sin A} = \frac{8}{2} = 4 \text{ cm} $$ The diameter $$ D $$ of the circumcircle is given by: $$ D = 2R $$ Thus, substituting the value of $$ R $$: $$ D = 2 \times 4 = 8 \text{ cm} $$ Therefore, the diameter of the circumcircle of triangle $$ \triangle ABC $$ is $$ 8 \text{ cm} $$. The answer provided in the options seems to be incorrect as it states $$ 4 \text{ cm} $$. The correct diameter is $$ 8 \text{ cm} $$.

The drameter length of the circumcircle of \( \triangle \mathrm{ABC} \) in which a \( =8 \mathrm{sin} \mathrm{Acm} \) equals \( \begin{array}{llll}\text { a) } 4 & \mathrm{~cm}\end{array} \)

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The drameter length of the circumcircle of \( \triangle \mathrm{ABC} \) in which a \( =8 \mathrm{sin} \mathrm{Acm} \) equals \( \begin{array}{llll}\text { a) } 4 & \mathrm{~cm}\end{array} \)

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