Question

Cho tam giac ABC co \(\widehat{A}\)=80, \(\widehat{B}\)=60. Tren tia doi BC lay diem M, tren tia doi cua CB lay diem N sao cho BM=BA, CN=CA. Tinh so do cac goc cua tam giac AMN

Answer 1

ta có góc C = 180-80-60=40⁰

Ta có :

\(\widehat{ACN}+\widehat{ACB}=180^0\\ \Rightarrow\widehat{ACN}=180^0-40^0=140^0\)

Ta lại có : CA=CN

=> tam giác ACN cân

=> \(\widehat{CAN}=\widehat{N}\)

\(\Rightarrow\widehat{CAN}+\widehat{N}=180^0-140^0=40^0\\ \Rightarrow\widehat{CAN}=\widehat{N}=20^0\)

\(\widehat{ABM}+\widehat{B}=180^0\\ \Rightarrow\widehat{ABM}=180^0-60^0=120^0\)

Ta lại có :

BA=BM => tam giác ABM cân

=> \(\widehat{MAB}=\widehat{M}\\ \Rightarrow\widehat{MAB}+\widehat{M}=180^0-120^0=60^0\\ \Rightarrow\widehat{MAB}=\widehat{M}=30^0\)

\(\widehat{A}\) của tam giác AMN = \(20^0+30^0+80^0=130^0\)

Chúc bạn học tốt !!!