$\Delta$ HBC vuông tại H có:
$HB=BC.sinC=5\sqrt3 (cm)$
$HC= BC.cosC=5(cm)$
$sinHBC=\frac{HC}{BC}=\frac{1}{2}$
$\Rightarrow \widehat{HBC}=30^o$
$\Rightarrow \widehat{ABH}=70-30=40^o$
$\Delta$ AHB vuông tại H có:
$HB=AB.cosABH$
$\Rightarrow AB=\frac{HB}{cosABH}\approx 11,3(cm)$