a/ Xét tứ giác $ABCD$:
$\widehat A+\widehat B+\widehat C+\widehat D=360°$
hay $30 °+70 °+\widehat B+\widehat D=360 °$
$↔100 °+\widehat B+\widehat D=360 °$
$↔\widehat B+\widehat D=260 °$
$↔\widehat B=260 °-\widehat D$
Thay $\widehat B=260 °-\widehat D$ vào $3\widehat B=4\widehat D$
$3(260 °-\widehat D)=4\widehat D$
$↔780 °-3\widehat D=4\widehat D\\↔-7\widehat D=-780 °\\↔\widehat D≈111,4 °\\→\widehat B=148,6 °$
Vậy $\widehat B=148,6 °,\widehat D=111,4 °$
b/ Xét tứ giác $ABCD$:
$\widehat A+\widehat B+\widehat C+\widehat D=360 °$
hay $\widehat A+40 °+\widehat C+80 °=360 °$
$↔\widehat A+\widehat C+120 °=360 °$
$↔\widehat A+\widehat C=240 °$
$↔\widehat A=240 °-\widehat C$
Thay $\widehat A=240 °-\widehat C$ vào $2\widehat A=3\widehat C$
$2(240 °-\widehat C)=3\widehat C\\↔480 °-2\widehat C=3\widehat C\\↔-5\widehat C=-480 °\\↔\widehat C=96 °\\→\widehat A=144 °$
Vậy $\widehat A=144 °,\widehat C=96 °$
