Question

Cho hình xẽ bên, cho biết AC//FD và \(\widehat{CBE}=126^o\)

Tính : \(\widehat{FEB}\) , \(\widehat{BED}\)

Answer 1

Vì AC//DF => \(\widehat{CBE}=\widehat{FEB}=126^o\)

(2 góc so le trong)

Vì AC//DF

=> \(\widehat{CBE}+\widehat{BED}=180^o\)

mà \(\widehat{CBE}=126^o\)

=>\(126^o+\widehat{BED}=180^o\)

=>\(\widehat{BED}=180^o-126^o\)

=>\(\widehat{BED}=54^o\)

Vậy\(\widehat{FEB}=126^o;\widehat{BED}=54^o\)

Answer 2

Giải:

Vì AC//FD

=> \(\widehat{FEB}\)=\(\widehat{CBE}\)=126^o (2 góc so le trong)

=> \(\widehat{BED}\)+\(\widehat{CBE}\)=180^o (2 góc trong cùng phía)

= \(\widehat{BED}\)+126^o =180^o

=\(\widehat{CBE}\)= 180^o -126^o

^{=\(\widehat{CBE}\)} = 54^o

Vậy: \(\widehat{CBE}\) = 54^o

\(\widehat{FEB}\) = 126^o