Xét ΔABC có
\(\widehat{BAC}+\widehat{B}+\widehat{C}=180^0\)
=>\(\widehat{BAC}=180^0-70^0-30^0=80^0\)
AD là phân giác của góc BAC
=>\(\widehat{BAD}=\widehat{CAD}=\dfrac{80^0}{2}=40^0\)
Xét ΔADB có \(\widehat{ADB}+\widehat{BAD}+\widehat{B}=180^0\)
=>\(\widehat{ADB}=180^0-40^0-70^0=70^0\)
=>\(\widehat{HAD}=90^0-70^0=20^0\)