a/ Có Dx⊥a
Ey⊥a
=> Dx // Ey
b/ Kẻ Hz // Dx
Mà Dx // Ey (câu a)
=> Hz // Dx // Ey
*Có Dx // Hz
=> \(\widehat{DHz}=\widehat{D}=25^0\)
*Có Hz // Ey
=> \(\widehat{zHE}=\widehat{E}=40^0\)
\(\widehat{DHE}=\widehat{DHz}+\widehat{zHE}\)
= 250 + 400 = 650
c/ Có Dx // Ey
=> \(\widehat{xPQ}+\widehat{yQP}\) = 1800 (2 góc trong cùng phía)
=> \(\widehat{yQP}=180^0-\widehat{xPQ}\) = 1800 - 1100 = 700
a) Ta có:
\(\left\{{}\begin{matrix}Dx\perp a\left(gt\right)\\Ey\perp a\left(gt\right)\end{matrix}\right.\)
=> \(Dx\) // \(Ey\) (từ vuông góc đến song song).
b) Qua \(H\) ta kẻ \(HN\) sao cho \(HN\) // \(Dx.\)
=> \(\widehat{DHN}=\widehat{xDH}\) (vì 2 góc so le trong).
Mà \(\widehat{xDH}=25^0\left(gt\right)\)
=> \(\widehat{DHN}=25^0.\)
Vì \(Dx\) // \(Ey\left(cmt\right)\)
\(Dx\) // \(HN\) (do cách vẽ).
=> \(HN\) // \(Ey.\)
=> \(\widehat{NHE}=\widehat{HEy}\) (vì 2 gsc so le trong).
Mà \(\widehat{HEy}=40^0\left(gt\right)\)
=> \(\widehat{NHE}=40^0.\)
Ta có: \(\widehat{DHE}=\widehat{DHN}+\widehat{NHE}\)
=> \(\widehat{DHE}=25^0+40^0\)
=> \(\widehat{DHE}=65^0.\)
c) Vì \(Dx\) // \(Ey\left(cmt\right)\)
=> \(\widehat{xPQ}+\widehat{PQy}=180^0\) (vì 2 góc trong cùng phía).
=> \(110^0+\widehat{PQy}=180^0\)
=> \(\widehat{PQy}=180^0-110^0\)
=> \(\widehat{PQy}=70^0\)
Vậy \(\widehat{PQy}=70^0.\)
Chúc bạn học tốt!