Question

Answer 1

Ta có: \(\widehat{ABC}=\widehat{BCx}=40^0\) (2 góc sole trong)

          \(\widehat{EDC}=\widehat{DCx}=65^0\) (2 góc sole trong)

mà \(\widehat{BCx}+\widehat{DCx}=\widehat{BCD}\)

\(\Rightarrow\) \(\widehat{BCD}=65^0+40^0=105^0\)

 

Answer 2

+)\(\widehat{E_1}=\widehat{A_1}=90^o\)

=>EA⊥AB

EA⊥ED

=>AB//ED

+)Kẻ Cx//AB

=>Cx//ED(vì cùng song song AB)

=>\(\widehat{DCx}=\widehat{FDC}=65^o\) _{(so le trong)}

+)Cx//AB

\(=>\widehat{BCx}=\widehat{ABC}=40^o\)_{(so le trong)}

+)\(\widehat{BCD}=\widehat{DCx}+\widehat{BCx}=65^o+40^o=105^o\)

Answer 3

vì Cx//AB

=> góc ABC=BCx=40 độ (2 góc sole trong)

vì Cx//DE

=> góc FDC=DCx=65 độ (2 góc sole trong)

mà góc BCD = BCx + DCx

góc BCD = 40 + 65

góc BCD = 105 độ