Question

Giải gấp giúp mình ạ

Answer 1

Bài 5:

Theo hình vẽ, ta có: Bx//ab//Dy

Ta có: Bx//ab

=>\(\widehat{xBC}=\widehat{bCB}\)(hai góc so le trong)

mà \(\widehat{xBC}=40^0\)

nên \(\widehat{bCB}=40^0\)

Ta có: ab//Dy

=>\(\widehat{bCD}=\widehat{yDC}\)(hai góc so le trong)

mà \(\widehat{yDC}=25^0\)

nên \(\widehat{bCD}=25^0\)

\(\widehat{BCD}=\widehat{bCB}+\widehat{bCD}\)

\(=25^0+40^0\)

\(=65^0\)

Bài 6:

Theo hình vẽ, ta có: Ex//ab//Gy

Ta có: Ex//ab

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

=>\(\widehat{bDE}+140^0=180^0\)

=>\(\widehat{bDE}=180^0-140^0=40^0\)

ab//Gy

=>\(\widehat{bDG}=\widehat{yGD}\)(hai góc so le trong)

mà \(\widehat{yGD}=30^0\)

nên \(\widehat{bDG}=30^0\)

\(\widehat{EDG}=\widehat{bDE}+\widehat{bDG}\)

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

\(=70^0\)