Question

Bài 9: Cho  có , Đường thẳng AD // BC.

Tính số đó .

Answer 1

Xét ΔABC có \(\widehat{A_1}+\widehat{B}+\widehat{C}=180^0\)

=>\(\widehat{A_1}=180^0-70^0-30^0=80^0\)

BC//AD

=>\(\widehat{A_2}=\widehat{C}\)(hai góc so le trong)

=>\(\widehat{A_2}=30^0\)

\(\widehat{DAB}=\widehat{A_1}+\widehat{A_2}=80^0+30^0=110^0\)

Answer 2

Ta có:

\(\widehat{A_1}=180-70-30=80^o\)(tính chất 3 góc trong tam giác)

\(\widehat{A_2}=\widehat{C}=30^o\)(cặp góc so le trong)

\(\widehat{BAD}=\widehat{A_1}+\widehat{A_2}=80+30=110^o\)