ΔABC=ΔMNP
=>\(\widehat{C}=\widehat{P}\)
mà \(\widehat{P}=50^0\)
nên \(\widehat{C}=50^0\)
Xét ΔABC có \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)
=>\(\widehat{A}+\widehat{B}+50^0=180^0\)
=>\(\widehat{A}+\widehat{B}=130^0\)
mà \(\widehat{A}-\widehat{B}=70^0\)
nên \(\widehat{A}=\dfrac{130^0+70^0}{2}=100^0\)