Question

Answer 1

Qua C, kẻ tia CM nằm giữa hai tia CB và CE sao cho CM//BA//DE

CM//BA

=>\(\hat{ABC}=\hat{BCM}\) (hai góc so le trong)

=>\(\hat{BCM}=40^0\)

CM//ED
=>\(\hat{MCE}=\hat{CED}\) (hai góc so le trong)

=>\(\hat{MCE}=30^0\)

Ta có: tia CM nằm giữa hai tia CB và CE

=>\(\hat{BCE}=\hat{BCM}+\hat{MCE}\)

\(=40^0+30^0=70^0\)