Bài 3:
1: m\(\perp\)r
n\(\perp\)r
Do đó: m//n
2: ta có: m//n
=>\(\widehat{H_3}=\widehat{I_2}\)(hai góc so le trong)
=>\(\widehat{I_2}=50^0\)
Ta có: \(\widehat{I_2}+\widehat{I_1}=180^0\)(hai góc kề bù)
=>\(\widehat{I_1}+50^0=180^0\)
=>\(\widehat{I_1}=130^0\)
Bài 4:
1: m//n
x\(\perp\)m
Do đó: x\(\perp\)n
2: ta có: m//n
=>\(\widehat{A_1}=\widehat{B_1}\)(hai góc so le trong)
=>\(\widehat{A_1}=60^0\)
Ta có: \(\widehat{A_2}=\widehat{A_1}\)(hai góc đối đỉnh)
mà \(\widehat{A_1}=60^0\)
nên \(\widehat{A_2}=60^0\)
Bài 5:
1: a\(\perp\)CD
b\(\perp\)CD
Do đó: a//b
2: a//b
=>\(\widehat{A_2}=\widehat{ABD}\)(hai góc so le trong)
=>\(\widehat{A_2}=45^0\)
Ta có: \(\widehat{A_2}+\widehat{A_1}=180^0\)(hai góc kề bù)
=>\(\widehat{A_1}+45^0=180^0\)
=>\(\widehat{A_1}=135^0\)
Ta có: \(\widehat{A_3}=\widehat{A_1}\)(hai góc đối đỉnh)
mà \(\widehat{A_1}=135^0\)
nên \(\widehat{A_3}=135^0\)
