Câu hỏi:

Ta có: $\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{6 + (\sqrt{3}+1)^2 - 4}{2\sqrt{6}(\sqrt{3}+1)} = \frac{6 + 3 + 2\sqrt{3} + 1 - 4}{2\sqrt{6}(\sqrt{3}+1)} = \frac{6 + 2\sqrt{3}}{2\sqrt{6}(\sqrt{3}+1)} = \frac{2\sqrt{3}(\sqrt{3}+1)}{2\sqrt{6}(\sqrt{3}+1)} = \frac{\sqrt{3}}{\sqrt{6}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$
Suy ra $A = 45^\circ$.
Áp dụng định lý sin, ta có: $\frac{a}{\sin A} = 2R \Rightarrow R = \frac{a}{2\sin A} = \frac{2}{2\sin 45^\circ} = \frac{1}{\frac{\sqrt{2}}{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.