Do $AB//CD$ $(gt)$
nên $\widehat{B} + \widehat{C} = 180^o$ (hai góc trong cùng phía)
mà $\widehat{B} - \widehat{C} = 60^o$
⇒ $\widehat{B} = \dfrac{180 + 60}{2} = 120^o$
⇒ $\widehat{C} = 120 - 60 = 60^o$
Ta có: $\widehat{D} = \dfrac{4}{5}\widehat{C}$
⇒ $\widehat{D} = \dfrac{4}{5}.60 = 48^o$
⇒ $\widehat{A} = 180 - \widehat{D} = 180 - 48 = 132^o$