Giải thích các bước giải:
Ta có :
$\widehat{ADB}=\dfrac{3}2\widehat{C}\to \widehat{C}=\dfrac 23\widehat{ADB}=\dfrac{160^o}3$
$\to \widehat{DAC}+\widehat{ACD}=\widehat{ADB}\to\widehat{DAC}=\dfrac{80^o}3$
$\to\widehat{BAC}=2\widehat{DAC}=\dfrac{160^o}3$
$\to\widehat{ABC}=180^o-\widehat{BAC}-\widehat{ACB}=\dfrac{220^o}3$