Cho \(\Delta ABC\) có \(AB = 3,AC = 4\;A\), diện tích \(S = 3\sqrt 3 \). Khi đó:

a) \(B{C^2} = A{B^2} + A{C^2} - 2AB \cdot AC \cdot \cos A\)

b) \(\sin A =  - \frac{{\sqrt 3 }}{2}\)

c) \(\cos A = \frac{1}{2}\)

d) \(\cos A =  - \frac{1}{2}\)

Ta có: \(S = \frac{1}{2}AB \cdot AC \cdot \sin A \Rightarrow \sin A = \frac{{2S}}{{AB \cdot AC}} = \frac{{2 \cdot 3\sqrt 3 }}{{3 \cdot 4}} = \frac{{\sqrt 3 }}{2}\). \({\cos ^2}A = 1 - {\sin ^2}A = 1 - \frac{3}{4} = \frac{1}{4} \Rightarrow \left[ {\begin{array}{*{20}{l}}{\cos A = \frac{1}{2}}\\{\cos A =  - \frac{1}{2}}\end{array};B{C^2} = A{B^2} + A{C^2} - 2AB \cdot AC \cdot \cos A} \right.\).

Với \(\cos A = \frac{1}{2}:B{C^2} = {3^2} + {4^2} - 2 \cdot 3 \cdot 4 \cdot \frac{1}{2} = 13 \Rightarrow BC = \sqrt {13} \).

Với \(\cos A =  - \frac{1}{2}:B{C^2} = {3^2} + {4^2} - 2 \cdot 3 \cdot 4 \cdot \left( { - \frac{1}{2}} \right) = 37 \Rightarrow BC = \sqrt {37} \).