Question

Answer 1

a: Ta có: \(\widehat{ADb}+\widehat{ADC}=180^0\)(hai góc kề bù)

=>\(\widehat{ADC}=180^0-120^0=60^0\)

Ta có: \(\widehat{ADC}=\widehat{A_1}\left(=60^0\right)\)

mà hai góc này là hai góc ở vị trí so le trong

nên BA//CD

=>\(\widehat{ABC}+\widehat{BCD}=180^0\)

=>\(x+80^0=180^0\)

=>\(x=100^0\)

b: Qua B, kẻ tia BD nằm giữa hai tia BA và BC sao cho BD//Ax//Cy

BD//Cy

=>\(\widehat{DBC}+\widehat{BCy}=180^0\)(hai góc trong cùng phía)

=>\(\widehat{DBC}=180^0-125^0=55^0\)

BD//Ax

=>\(\widehat{xAB}+\widehat{ABD}=180^0\)(hai góc trong cùng phía)

=>\(\widehat{ABD}=180^0-130^0=50^0\)

\(\widehat{ABC}=\widehat{ABD}+\widehat{CBD}=50^0+55^0=105^0\)