Bài giải:

Vì AB // CD nên $\left\{ \begin{align}& \widehat{A}+\widehat{D}={{180}^{0}} \\ & \widehat{B}+\widehat{C}={{180}^{0}} \\ \end{align} \right.$ (hai góc trong cùng phía)

Mà: $\left\{ \begin{align}& \widehat{A}+\widehat{D}={{180}^{0}} \\ & \widehat{A}-\widehat{D}={{20}^{0}} \\ \end{align} \right.$

$\Rightarrow \widehat{A}=\frac{{{180}^{0}}+{{20}^{0}}}{2}={{100}^{0}}$

$\left\{ \begin{align}& \widehat{B}+\widehat{C}={{180}^{0}} \\ & \widehat{B}=\frac{3}{2}\widehat{C} \\ \end{align} \right.$

$\Rightarrow \frac{3}{2}\widehat{C}+\widehat{C}={{180}^{0}}=\frac{5}{2}\widehat{C}$

$\Rightarrow \widehat{C}={{72}^{0}}$

Suy ra: $\widehat{B}=\frac{3}{2}{{.72}^{0}}={{108}^{0}}$

Vậy $\widehat{A}={{100}^{0}}$ ; $\widehat{B}={{108}^{0}}$