In triangle ABC,if b=2, c=sqrt{3},and A=30^{c | vedclass

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(D) Using the Law of Cosines,$a^2 = b^2 + c^2 - 2bc \cos A$. Substituting the values: $a^2 = 2^2 + (\sqrt{3})^2 - 2(2)(\sqrt{3}) \cos 30^{\circ} = 4 +

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$\frac{\sqrt{3}-1}{2}$

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(D) Using the Law of Cosines,$a^2 = b^2 + c^2 - 2bc \cos A$. Substituting the values: $a^2 = 2^2 + (\sqrt{3})^2 - 2(2)(\sqrt{3}) \cos 30^{\circ} = 4 + 3 - 4\sqrt{3} 	imes \frac{\sqrt{3}}{2} = 7 - 6 = 1$. Thus,$a = 1$. The area of the triangle $\Delta = \frac{1}{2} bc \sin A = \frac{1}{2} (2)(\sqrt{3}) \sin 30^{\circ} = \sqrt{3} 	imes \frac{1}{2} = \frac{\sqrt{3}}{2}$. The semi-perimeter $s = \frac{a+b+c}{2} = \frac{1+2+\sqrt{3}}{2} = \frac{3+\sqrt{3}}{2}$. The inradius $r = \frac{\Delta}{s} = \frac{\sqrt{3}/2}{(3+\sqrt{3})/2} = \frac{\sqrt{3}}{3+\sqrt{3}}$. Rationalizing the denominator: $r = \frac{\sqrt{3}(3-\sqrt{3})}{(3+\sqrt{3})(3-\sqrt{3})} = \frac{3\sqrt{3}-3}{9-3} = \frac{3(\sqrt{3}-1)}{6} = \frac{\sqrt{3}-1}{2}$.

In $\triangle ABC$,if $b=2, c=\sqrt{3}$,and $A=30^{\circ}$,then its inradius $r=$

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