Dựng đường cao $AH$ vuông góc $BC$
Xét $ΔABH$ vuông tại $H$:
$·\cos B=\dfrac{BH}{AB}$ hay $\cos 60^\circ=\dfrac{BH}{3}$
$↔\dfrac{1}{2}=\dfrac{BH}{3}\\↔BH=1,5cm\\→CH=5-1,5=3,5cm$
$·\sin B=\dfrac{AH}{AB}$ hay $\sin 60^\circ=\dfrac{AH}{3}$
$↔\dfrac{\sqrt 3}{2}=\dfrac{AH}{3}\\↔AH=\dfrac{3\sqrt 3}{2}cm$
Xét $ΔACH$ vuông tại $H$:
$\tan C=\dfrac{AH}{CH}$ hay $\tan C=\dfrac{\dfrac{3\sqrt 3}{2}}{3,5}$
$↔\tan C=\dfrac{3\sqrt 3}{7}\\↔\widehat C≈36,6^\circ$
$·\sin C=\dfrac{AH}{AC}$ hay $\sin 36,6^\circ=\dfrac{\dfrac{3\sqrt 3}{2}}{AC}$
$↔AC=\dfrac{\dfrac{3\sqrt 3}{2}}{\sin 36,6^\circ}\\↔AC≈4,36cm$
Xét $ΔABC$:
$\widehat A+\widehat B+\widehat C=180^\circ$
hay $\widehat A+60^\circ+36,6^\circ=180^\circ$
$↔\widehat A=83,4^\circ$
Vậy $\widehat A=83,4^\circ; \,\widehat C=36,6^\circ;\, AC=4,36cm$
