\(\widehat{C}=\frac{5}{6}\widehat{B}\Rightarrow\widehat{\frac{C}{5}}=\frac{\widehat{B}}{6}\)
Mà \(\Delta ABC=\Delta DEP\)\(\Rightarrow\widehat{E}=\widehat{B}\) và \(\widehat{C}=\widehat{F}\) ( Hai cặp góc tương ứng )
Mà \(\widehat{E}-\widehat{F}=10^o\Rightarrow\widehat{B}-\widehat{C}=10^o\)
Áp dụng tính chất dãy tỉ số bằng nhau , ta có :
\(\frac{\widehat{B}}{6}=\frac{\widehat{C}}{5}=\frac{\widehat{B}-\widehat{C}}{6-5}=\frac{10^o}{1}=10^o\)
Có : \(\frac{\widehat{B}}{6}=10^o\Rightarrow\widehat{B}=10^o.6=60^o\)
\(\frac{\widehat{C}}{5}=10^o\Rightarrow\widehat{C}=10^o.5=50^o\)
Mà \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\Rightarrow\widehat{A}+60^o+50^o=180^o\)
\(\Rightarrow\widehat{A}=180^o-60^o-50^o=70^o\)
Lại có : \(\widehat{A}=\widehat{D}=70^o;\widehat{B}=\widehat{E}=60^o;\widehat{C}=\widehat{F}=50^o\)
Kết luận ...
Ta có \(\Delta ABC=\Delta DEF\)
=> \(\widehat{A}=\widehat{D};\widehat{C}=\widehat{F};\widehat{B}=\widehat{E}\)
=> \(\widehat{C}=\frac{5}{6}\widehat{B}\Rightarrow\widehat{F}=\frac{5}{6}\widehat{E}\)
Ta lại có: \(\widehat{E}-\widehat{F}=10^o\)=> \(\widehat{E}-\frac{5}{6}\widehat{E}=10^o\)=> \(\widehat{E}=60^o\)
=> \(\widehat{F}=\frac{5}{6}.60^o=50^o\)=> \(\widehat{D}=180^o-\widehat{E}-\widehat{F}=180^o-50^o-60^o=70^o\)
=> \(\widehat{A}=\widehat{D}=70^o;\widehat{C}=\widehat{F}=50^o;\widehat{B}=\widehat{E}=60^o\)
