Lời giải:
a)
$\widehat{ADC}+\widehat{ADB}=\widehat{BDC}=180^0(1)$
$\widehat{ADC}-\widehat{ADB}=(\widehat{BAD}+\widehat{B})-(\widehat{CAD}+\widehat{C})$
$=\widehat{B}-\widehat{C}=30^0(2)$
Từ $(1);(2)$ suy ra:
$\widehat{ADC}=\frac{180^0+30^0}{2}=105^0$
$\widehat{ADB}=\frac{180^0-30^0}{2}=75^0$
b)
Vì tam giac $HAD$ vuông tại $H$ nên:
$\widehat{HAD}=90^0-\widehat{ADH}=90^0-\widehat{ADB}=90^0-75^0=15^0$