Cho \(\Delta ABC=\Delta MNP\), \(\widehat{A}=30^0\), \(\widehat{P}=60^0\). So sánh các góc \(N\), \(M\), \(P\)?
\(\widehat{N}=\widehat{P}>\widehat{M}\).\(\widehat{N}>\widehat{P}=\widehat{M}\).\(\widehat{N}>\widehat{P}>\widehat{M}\).\(\widehat{N}< \widehat{P}< \widehat{M}\).Hướng dẫn giải:Do \(\Delta ABC=\Delta MNP\) nên \(\widehat{A}=\widehat{M}=30^0\), \(\widehat{B}=\widehat{N}\), \(\widehat{C}=\widehat{P}=60^0\)
Xét trong tam giác \(MNP\) có \(\widehat{M}+\widehat{N}+\widehat{P}=180^0\) \(\Rightarrow\widehat{N}=180^0-\widehat{M}-\widehat{P}=180^0-60^0-30^0=90^0\)
Có \(90^0>60^0>30^0\) nên \(\widehat{N}>\widehat{P}>\widehat{M}\)
