Cho Δ ABC có BC = √ 6 , CA = 2 , AB = 1 + √ 3 . Khi đó: a) ˆ A = 30 độ

cosA=b2+c2−a22bc=12⇒A^=60°;cosB=a2+c2−b22ac=22⇒B^=45°

\(S = \frac{1}{2}bc\sin A = \frac{{3 + \sqrt 3 }}{2};S = \frac{1}{2}BC \cdot AH \Rightarrow AH = \frac{{2S}}{{BC}};S = \frac{{abc}}{{4R}} \Rightarrow R = \frac{{abc}}{{4S}} = \sqrt 2 .\)