Cho tam giác ABC có

- Hướng dẫn giải

a) Ta có $\widehat{A}=2.\widehat{B}$; $\widehat{B}=3.\widehat{C}\Rightarrow \widehat{A}=6\widehat{C}$.

$\Delta ABC$ có $\widehat{A}+\widehat{B}+\widehat{C}=180°\Rightarrow 6.\widehat{C}+3\widehat{C}+\widehat{C}=180°$

$\Rightarrow \widehat{C}=18°;\widehat{B}=54°;\widehat{A}=108°$.

b) Ta có $\widehat{ACx}+\widehat{C_1}=180°$(hai góc kề bù)

Ta có: $\widehat{ACx}+18°=180°\Rightarrow \widehat{ACx}=162°$.

$\Delta BCE$ có $\widehat{E}+\widehat{B}+\widehat{BCE}=180°;\widehat{E}+54°+18°+81°=180°\Rightarrow \widehat{E}=27°$ hay $\widehat{AEC}=27°$.