Xét ΔANC và ΔAMB có:
\(\widehat{AMB}=\widehat{ANC}=90^o\)
\(\widehat{NAC}\) chung
\(\Rightarrow\text{Δ}ANC\sim\text{ΔAMB}\left(g.g\right)\)
\(\Rightarrow\dfrac{AM}{AN}=\dfrac{AB}{AC}\)
\(\Rightarrow\dfrac{40}{40+MN}=\dfrac{60}{90+60}\)
\(\Rightarrow60\left(40+MN\right)=40\cdot150\)
\(\Rightarrow2400+60MN=6000\)
\(\Rightarrow60MN=3600\)
\(\Rightarrow MN=60\left(cm\right)\)
Vậy: ...
Có \(MB // NC\) ( 2 bên bờ ao cá ), theo định lí Thales có :
`{AM}/{MN} = {AB}/{BC}`
\( \Leftrightarrow \frac{40}{MN}=\frac{60}{90}\\ \Leftrightarrow MN . 60 =40 . 90 \\ \Leftrightarrow MN . 60= 3600\\ \Leftrightarrow MN = 3600:60\\ \Leftrightarrow MN=60(m)\)
Vậy \(MN= 60(m)\)
