Xét ΔABH có \(\widehat{A}+\widehat{H}+\widehat{ABH}=180^0\)
=>\(\widehat{ABH}=180^0-40^0-70^0=70^0\)
Sửa đề; BH=3cm
Xét ΔBHA có \(\dfrac{BH}{sinA}=\dfrac{AH}{sinABH}=\dfrac{BA}{sinAHB}\)
=>\(\dfrac{3}{sin40}=\dfrac{AH}{sin70}=\dfrac{BA}{sin70}\)
=>\(AH=BA=3\cdot\dfrac{sin70}{sin40}\simeq4,39\left(cm\right)\)