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In $$\triangle ABC$$, $$\angle ACB = 30^\circ$$, $$AB = 5$$ cm. Then the circumradius of $$\triangle ABC$$ is
Correct Answer: 5
By the extended sine rule, $$\frac{AB}{\sin C} = 2R$$ where R is the circumradius. Here $$\frac{5}{\sin 30^\circ} = \frac{5}{\frac{1}{2}} = 10 = 2R$$. Hence $$R = 5$$ cm.
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