a) $xx'//yy'$
$→\widehat{yBA}+\widehat{A_1}=180^o$ (trong cùng phía bù nhau)
mà $\widehat{yBA}=140^o$
$→\widehat{A_1}=180^o-140^o=40^o$
$xx'//yy'$
$→\widehat{y'CA}=\widehat{A_3}=70^o$ (so le trong)
Vì $\widehat{A_1}+\widehat{A_2}+\widehat{A_3}=180^o$
mà $\widehat{A_1}=40^o, \widehat{A_3}=70^o$
$→\widehat{A_2}=180^o-40^o-70^o=70^o$
b) Vì $\widehat{yBA}+\widehat{ABC}=180^o$ (kề bù)
mà $\widehat{yBA}=140^o$
$→\widehat{ABC}=180^o-140^o=40^o$
mà $Bn$ là đường phân giác $\widehat{ABC}$
$→\widehat{nBC}=\dfrac{\widehat{ABC}}{2}=\dfrac{40^o}{2}=20^o$
Xét $ΔBKC$:
$\widehat{CBK}+\widehat{BCK}+\widehat{BKC}=180^o$ (tổng 3 góc trong 1 Δ)
mà $\widehat{CBK}=20^o, \widehat{BCK}=70^o$
$→\widehat{BKC}=180^o-20^o-70^o=90^o$
mà $\widehat{BKC}$ và $\widehat{CKn}$ là 2 góc kề bù
$→\widehat{CKn}=180^o-90^o=90^o$
