Ta có:
\(5\widehat{B}\) bù với \(\widehat{A}\Rightarrow5\widehat{B}+\widehat{A}=180^0\)
\(\Rightarrow\widehat{A}=180^0-5\widehat{B}\) (1)
\(2\widehat{B}\) phụ với \(\widehat{A}\Rightarrow2\widehat{B}+\widehat{A}=90^0\) (2)
Thay (1) vào (2), ta có:
\(2\widehat{B}+180^0-5\widehat{B}=90^0\)
\(3\widehat{B}=90^0\)
\(\widehat{B}=30^0\) thay vào (1) ta có:
\(\widehat{A}=180^0-5.30^0=30^0\)
Vậy \(\widehat{A}=\widehat{B}\)