5.16
a) Ta có: c⊥a(gt)
c⊥b(gt)
Do đó: a//b(Định lí 1 từ vuông góc tới song song)
b) Ta có: a//b(cmt)
nên \(\widehat{CDB}=\widehat{C_1}\)(hai góc so le trong)
mà \(\widehat{CDB}=60^0\)(gt)
nên \(\widehat{C_1}=60^0\)
mà \(\widehat{C_3}=\widehat{C_1}\)(hai góc đối đỉnh)
nên \(\widehat{C_3}=60^0\)
Ta có: \(\widehat{C_2}+\widehat{C_3}=180^0\)(hai góc kề bù)
\(\Leftrightarrow\widehat{C_2}+60^0=180^0\)
hay \(\widehat{C_2}=120^0\)
mà \(\widehat{C_4}=\widehat{C_2}\)(hai góc đối đỉnh)
nên \(\widehat{C_4}=120^0\)
Vậy: \(\widehat{C_1}=60^0\); \(\widehat{C_2}=120^0\); \(\widehat{C_3}=60^0\); \(\widehat{C_4}=120^0\)
a, \(Dx\perp a,Ey\perp a=>Dx//Ey\)
b,nối DE \(=>\angle\left(EDM\right)+\angle\left(DEN\right)=180^o\)( góc trong cùng phía)
\(=>\angle\left(EDH\right)+\angle\left(DEH\right)=180-35-40=105^o\)
\(=>\angle\left(DHE\right)=180^o-105^0=75^0\)
c,\(\angle\left(PQy\right)+\angle\left(xPQ\right)=180^o\)(góc trong cùng phía)
\(=>\angle\left(PQy\right)=180-110=70^o\)
