Do a // b nên
.)\(\widehat{A_2}=\widehat{B_4}=55^0(so le trong)\)
.)\(\widehat{B_4}+\widehat{B_1}=180^0\)
\(\Leftrightarrow\)\(\widehat{B_1}=180^0-55^0=125^0\)
.)\(\widehat{B_1}=\widehat{B_3}=125^0(đối đỉnh)\)
.)\(\widehat{A_4}=\widehat{A_2}=55^0(đối đỉnh)\)
Lời giải:
Do $a\parallel b$ nên $\widehat{B_4}=\widehat{A_2}=55^0$ (hai góc so le trong)
$\widehat{B_1}+\widehat{B_4}=180^0$
$\Rightarrow \widehat{B_1}=180^0-\widehat{B_4}=180^0-55^0=125^0$
$\widehat{B_3}=\widehat{B_1}=125^0$ (hai góc đối đỉnh)
$\widehat{A_4}=\widehat{A_2}=55^0$ (hai góc đối đỉnh)