Question

Answer 1

a: Ta có: MN//EF

ME\(\perp\)EF

Do đó: MN\(\perp\)ME

b: Ta có: \(\widehat{N_1}+\widehat{N_3}=180^0\)(hai góc kề bù)

=>\(\widehat{N_3}=180^0-110^0=70^0\)

c: MN//EF

=>\(\widehat{N_1}=\widehat{F_2}\)(hai góc đồng vị)

=>\(\widehat{F_2}=110^0\)

Ta có: \(\widehat{F_2}+\widehat{F_4}=180^0\)(hai góc kề bù)

=>\(\widehat{F_4}=180^0-110^0=70^0\)

Ta có: \(\widehat{F_4}=\widehat{F_3}\)(hai góc đối đỉnh)

mà \(\widehat{F_4}=70^0\)

nên \(\widehat{F_3}=70^0\)