Giải thích các bước giải:

Kẻ $CD\perp AB$

Ta có:

$\sin A=\dfrac{CD}{AC}$

$\to CD=AC\sin A=12\sin60^o= 6\sqrt3$

$\to AD=\sqrt{AC^2-CD^2}=\sqrt{12^2-(6\sqrt3)^2}=6$

$\to BD=AB-AD=2$

$\to BC=\sqrt{CD^2+BD^2}=\sqrt{(6\sqrt3)^2+2^2}=4\sqrt{7}$

Ta có:

$\tan B=\dfrac{CD}{BD}=\dfrac{6\sqrt3}2=3\sqrt3$

$\to \hat B\approx  79^o$

$\to \hat C=180^o-\hat A-\hat B= 41^o$