Ta có:
$\widehat{A}+\widehat{C}=180^o$
$↔ 3\widehat{A}+3\widehat{C}=540^o$
$↔ 2\widehat{C}+3\widehat{C}=540^o$
$↔ 5\widehat{C}=540^o$
$↔ \widehat{C}=108^o$
$→ \widehat{A}=180^o-\widehat{C}=72^o$
$\widehat{B}=(210^o-\widehat{A}):3=46^o$
$\widehat{D}=180^o-\widehat{B}=134^o$