Cách 1:
Ta có:
\(\widehat{BEA}+\widehat{AEC}=180^o\) (hai góc kề bù)
\(\Rightarrow\widehat{BEA}=180^o-80^o=100^o\)
Xét tam giác ABE ta có:
\(\widehat{BAE}+\widehat{BEA}+\widehat{ABE}=180^o\) (tổng 3 góc trong tam giác)
\(\Rightarrow\widehat{BAE}=180^o-60^o-100^o=20^o\)
Cách 2:
Ta có:
\(\widehat{ACE}+\widehat{AC..}=180^o\)
\(\Rightarrow\widehat{ACE}=180^o-145^o=35^o\)
Xét tam giác ABC ta có:
\(\widehat{ACE}+\widehat{ABC}+\widehat{BAC}=180^o\) (tổng 3 goc của tam giác)
\(\Rightarrow\widehat{BAC}=180^o-35^o-60^o=85^o\)
Xét tam giác AEC ta có:
\(\widehat{AEC}+\widehat{EAC}+\widehat{ACE}=180^o\)
\(\Rightarrow\widehat{EAC}=180^o-35^o-80^o=65^o\)
Mà: \(\widehat{EAC}+\widehat{BAE}=\widehat{BAC}\)
\(\Rightarrow\widehat{BAE}=\widehat{BAC}-\widehat{EAC}=85^o-65^o=20^o\)
