a/ Kẻ đường cao $CD$ ứng $AB$
Xét $ΔBCD$ vuông tại $D$:
$·\,\,\cos B=\dfrac{BD}{BC}$ hay $\cos 60^\circ=\dfrac{BD}{12}$
$↔\dfrac{1}{2}=\dfrac{BD}{12}\\→BD=6cm$
$·\,\,\sin B=\dfrac{CD}{BC}$ hay $\sin 60^\circ=\dfrac{CD}{12}$
$↔\dfrac{\sqrt 3}{2}=\dfrac{CD}{12}\\↔CD=6\sqrt 3$
Xét $ΔBCD$:
$\widehat B+\widehat{BCD}=90^\circ$ (vì $ΔBCD$ vuông tại $D$)
hay $60^\circ+\widehat{BCD}=90^\circ$
$→\widehat{BCD}=30^\circ$
$→\widehat{ACD}=\widehat C-\widehat{BCD}$
hay $\widehat{ACD}=40^\circ-30^\circ$
$→\widehat{ACD}=10^\circ$
Xét $ΔACD$ vuông tại $D$:
$\tan ACD=\dfrac{AD}{CD}$ hay $\tan 10^\circ=\dfrac{AD}{6\sqrt 3}$
$↔AD=\tan 10^\circ.6\sqrt 3≈1,83(cm)$
Ta có: $AB=BD+AD=6+1,83=7,83(cm)$
Xét $ΔABH$ vuông tại $H$:
$\sin B=\dfrac{AH}{AB}$ hay $\sin 60^\circ=\dfrac{AH}{7,83}$
$↔\dfrac{\sqrt 3}{2}=\dfrac{AH}{7,83}\\↔AH≈6,78(cm)$
Vậy $AH≈6,78$
b/ $S_{ΔABC}=\dfrac{1}{2}.AH.BC=\dfrac{1}{2}.6,78.12=40,68(cm^2)$
Vậy $S_{ΔABC}=40,68cm^2$
