Question

Answer 1

a: ta có: Ex\(\perp\)m

Fy\(\perp\)m

Do đó: Ex//Fy

Ta có: Ex//Fy

Cz//Ex

Do đó: Cz//Fy

b: Ta có: Cz//Ex

=>\(\widehat{zCE}=\widehat{xEC}\)(hai góc so le trong)

mà \(\widehat{xEC}=30^0\)

nên \(\widehat{zCE}=30^0\)

Ta có: \(\widehat{zCE}+\widehat{zCF}=\widehat{ECF}\)

=>\(\widehat{zCF}+30^0=110^0\)

=>\(\widehat{zCF}=80^0\)

Cz//Fy

=>\(\widehat{zCF}=\widehat{yFC}\)(hai góc so le trong)

mà \(\widehat{zCF}=80^0\)

nên \(\widehat{yFC}=80^0\)

c: ΔCAE vuông tại A

=>\(\widehat{ACE}+\widehat{AEC}=90^0\)

=>\(\widehat{ACE}+30^0=90^0\)

=>\(\widehat{ACE}=60^0\)