Giả sử tam giác \(ABC\) có \(\widehat A = 90^\circ ,\widehat C = 60^\circ ,AC = 3\).
Ta có: \(BC = \dfrac{{AC}}{{\cos 60^\circ }} = \dfrac{3}{{{1 \over 2}}} = 6\)
\(\sin 60^\circ = \sin \widehat C = \dfrac{{AB}}{{BC}}\)
Suy ra: \(AB = BC.\sin 60^\circ = 6. \dfrac{{\sqrt 3 }}{2} = 3\sqrt 3. \)
