Đáp án: $\widehat{PQR}=80^o$
Giải thích các bước giải:
Kẻ tia $PC//Ax, RD//By, QE//Ax$
Vì $Ax//By\to PC//RD//QE$
Ta có : $Ax//PC//QE$
$\widehat{APC}=\widehat{xAP}=30^o$ (so le trong)
$\to\widehat{CPQ}=\widehat{APQ}-\widehat{APC}=40^o$
$\to \widehat{PQE}=\widehat{CPQ}=40^o$ (so le trong)
Lại có: $QE//RD//BY$
$\to \widehat{DRB}=\widehat{RBY}=40^o$ (so le trong)
$\widehat{QRD}=\widehat{QRB}-\widehat{DRB}=40^o$
$\to \widehat{EQR}=\widehat{QRD}=40^o$ (so le trong)
$\to \widehat{PQR}=\widehat{PQE}+\widehat{EQR}=80^o$
