Bài 1.$\triangle BGC$ có $BG=4, CG=6 , \widehat{BGC}=120$
$BC^2=BG^2+CG^2-BG.CG.2cos120=4^2+6^2+4.6=76$
$\Rightarrow BC=\sqrt{76}=2\sqrt{19}$
$\triangle MGC$ có $MG=2 , CG=6 , \widehat{MGC}=60$
$\Rightarrow MC^2=MG^2+CG^2-MG.CG.2cos60=2^2+6^2-2.6=28$
$\Rightarrow AC=2MC=4\sqrt7$
$\triangle NGB$ có $NG=3, BG=4 , \widehat{NGB}=60$
$\Rightarrow NB^2=NG^2+BG^2-NG.BG.2cos60=3^2+4^2-3.4=13$
$\Rightarrow AB=2NB=2\sqrt{13}$