a/ $AB//CD$
$→\widehat B+\widehat C=180^\circ$ (2 góc trong cùng phía)
hay $140^\circ+x=180^\circ$
$↔x=180^\circ-140^\circ=40^\circ$
$AB//CD$
$→\widehat A+\widehat D=180^\circ$ (2 góc trong cùng phía)
hay $y+60^\circ=180^\circ$
$↔y=180^\circ-60^\circ=120^\circ$
Vậy $x=40^\circ,y=120^\circ$
b/ Xét tứ giác $HGKI$:
$\widehat H+\widehat G+\widehat K+\widehat K=360^\circ$
hay $4x+3x+2x+x=360^\circ$
$↔10x=360^\circ\\↔x=36^\circ$
hay $\widehat I=36^\circ$
$→\begin{cases}\widehat{H}=4.36^\circ=144^\circ\\\widehat G=3.36^\circ=108^\circ\\\widehat K=2.36^\circ=72^\circ\end{cases}$
Vậy $\widehat H=144^\circ,\widehat G=108^\circ,\widehat K=72^\circ,\widehat I=36^\circ$
