a,
\(\Delta ABC=\Delta PQR\\ \Rightarrow\widehat{A}=\widehat{P}=50^o\\ \widehat{B}=\widehat{Q}\)
Xét \(ABC\) có
\(\widehat{A}+\widehat{B}+\widehat{C}=180^o\)
\(\Rightarrow\widehat{B}+\widehat{C}=130^o\\ \Rightarrow\widehat{B}=130^o-\widehat{C}\)
\(\widehat{B}-\widehat{C}=50^o\\ \Rightarrow130^o-2\widehat{C}=50^o\\ \Rightarrow\widehat{C}-40^o\\ \Rightarrow\widehat{B}=90^o=\widehat{Q}\)
\(\Rightarrow PQR\) là tam giác vuông
b, \(\Delta ABC=\Delta PQR\\ \Rightarrow\left\{{}\begin{matrix}AC=PR\\AB=PQ\\BC=QR\end{matrix}\right.\)