\(\sin\widehat{C}=\dfrac{AB}{AC}\Rightarrow AB=AC.sin\widehat{C}=5.sin60^0=\dfrac{5\sqrt{3}}{2}cm\)
\(\widehat{A}=\widehat{B}-\widehat{C}=90^0-60^0=30^0\)
\(BC=\sqrt{AC^2-AB^2}=\sqrt{5^5-\dfrac{5\sqrt{3}}{2}^2}=2,5cm\)
BC=AC.cosC=5.\(\dfrac{1}{2}\)=2.5 cm
BA=AC.sinC=5.\(\dfrac{\sqrt{3}}{2}\)=\(\dfrac{5\sqrt{3}}{2}\)cm
Góc C = 900-600=300