a) Ta có $3\widehat{A1}=2\widehat{A2}$ hay $\widehat{A2}=\frac{3}{2}\widehat{A1}$ và $\widehat{A1}+\widehat{A2}=180^{\circ}$ (hai góc kề bù) nên
$\widehat{A1}+\frac{3}{2}\widehat{A1}=\frac{5}{2}\widehat{A1}=180^{\circ}$ hay $\widehat{A1} =72^{\circ}$. Do đó$ \widehat{A2}=\frac{3}{2}72^{\circ}=108^{\circ}$
Từ đó, ta tính được$ \widehat{A1}=\widehat{A3}=\widehat{B1}=\widehat{B3}=72^{\circ}, \widehat{A2}=\widehat{A4}\widehat{B2}=\widehat{B4}=108^{\circ}$