a) Trong $\Delta$ vuông $ACH$
$\widehat{HAC}=90^o-\widehat{HCA}=90^o-30^o=60^o$
Trong $\Delta $ vuông $ABH$
$\widehat{HAB}=90^o-\widehat{ABH}=90^o-70^o=20^o$
b) Theo tính chất tổng 3 góc trong một tam giác $\Delta ABC$
$\widehat A=180^o-\widehat B-\widehat C=180^o-70^o-30^o=80^o$
$AD$ là phân giác $\widehat A$
$\Rightarrow \widehat{DAB}=\widehat{DAC}=\dfrac{80^o}{2}=40^o$
Trong $\Delta ADC$
$\widehat{ADC}=180^o-\widehat{DAC}-\widehat{DCA}$
$=180^o-40^o-30^o=110^o$
$\widehat{ADB}=180^o-\widehat{ADC}=180^o-110^o=70^o$