[tex]\it m(\hat{B}) =\dfrac{\pi}{6}\ \ \ \ \ (1)
\\ \\ \\
m(\hat{C}) =\dfrac{\pi}{3} \ \ \ \ \ (2)
\\ \\ \\
(1),\ (2) \Rightarrow m(\hat{A}) =\dfrac{\pi}{2}[/tex]
Aplicăm teorema sinusurilor:
[tex]\it \dfrac{AC}{sinB} = \dfrac{BC}{sinA} \Rightarrow AC = \dfrac{BC sinB}{sinA} =\dfrac{4\cdot sin{\dfrac{\pi}{6}}}{sin\dfrac{\pi}{2}} = \dfrac{4\cdot\dfrac{1}{2}}{1} = 2[/tex]
