Ta có :
Góc BAC + góc ABC + góc ACB = 180o ( tổng 3 góc trong tam giác )
⇒80o + góc ABC + góc ACB = 180o
Góc ABC + góc ACB = 100o
⇒\(\frac{1}{2}\)góc ABC + \(\frac{1}{2}\) góc ACB = \(=\frac{100^0}{2}=50^0\)
⇒góc BCE + góc CBE = 50o
Lại xét ΔBEC:
Góc BEC + góc BCE + góc CBE = 180o
Góc BEC + 50o = 180o
Góc BEC = 130o
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