Răspuns :
[tex]\it cosB=\dfrac{16+3-7}{2\cdot4\sqrt3}=\dfrac{^{\sqrt3)}12}{\ 8\sqrt3}=\dfrac{12\sqrt3}{8\cdot3}=\dfrac{\ 12\sqrt3^{(12}}{24}=\dfrac{\sqrt3}{2}[/tex]
[tex]\it cosB=\dfrac{16+3-7}{2\cdot4\sqrt3}=\dfrac{^{\sqrt3)}12}{\ 8\sqrt3}=\dfrac{12\sqrt3}{8\cdot3}=\dfrac{\ 12\sqrt3^{(12}}{24}=\dfrac{\sqrt3}{2}[/tex]