TG ABC = TG DEF=> góc A  -=D = 60 độ

                                 góc B = E = 70 độ

                                 góc C = F = 180 - 60 -70 =50 độ

\(\widehat{D}=180^0-\widehat{E}-\widehat{F}=50^0=\widehat{A}\\ \left\{{}\begin{matrix}AB=DE\\\widehat{A}=\widehat{D}\\AC=DE\end{matrix}\right.\Rightarrow\Delta ABC=\Delta DEF\left(c.g.c\right)\)

Xét t/giác DEF có \(\widehat{D}+\widehat{E}+\widehat{F}=180^0\) (tổng 3 góc của 1 t/giác)

=> \(\widehat{D}=180^0-\widehat{E}-\widehat{F}=180^0-70^0-60^0=50^0\)

Xét t/giác ABC và t/giác DEF

có: AB = DE (gt)

   AC = DF (gt)

 \(\widehat{A}=\widehat{D}=50^0\)

=> t/giác ABC = t/giác DEF (c.g.c)

mình cần gấp ạ 

ΔABC = ΔDEF

⇒ \(\hept{\begin{cases}\widehat{A}=\widehat{D}=55^O\\\widehat{B}=\widehat{E}=75^O\\\widehat{C}=\widehat{F}\end{cases}} \)
Tổng ba góc trong tam giác bằng \(^{180^O}\)
  Hay 50^O + 75^O + \(\widehat{C}\)=180^O
=> C^{^} =F^{^} =50O

\(\Delta ABC=\Delta DEF\)nên \(\widehat{A}=\widehat{D}=55^0,\widehat{B}=\widehat{E}=75^0\)

Trong hai tam giác ABC và DEF ta có : 

=> \(\widehat{C}=180^0-\left[\widehat{A}+\widehat{B}\right]=180^0-\left[55^0+75^0\right]=50^0\)

\(\widehat{F}=180^0-\left[\widehat{D}+\widehat{E}\right]=180^0-\left[55^0+75^0\right]=50^0\)

\(\Leftrightarrow\widehat{C}=\widehat{F}=50^0\)

bạn tự vẽ hình giúp mik nha

vẽ đường cao EH (H\(\in\)DF)

ta có: \(\widehat{F}\)=180\(^o\)-\(\widehat{E}\)-\(\widehat{F}\)=180-70-60=50

EH=EF.sinF=30.sin50=22,98

sinD=\(\dfrac{EH}{ED}\)\(\Rightarrow\)ED=\(\dfrac{EH}{sinD}\)=\(\dfrac{22,98}{sin60}\)=26,54

DH=\(\sqrt{DE^2-EH^2}\)(pytago)=\(\sqrt{26,54^2-22,98^2}\)=13,28

HF=\(\sqrt{EF^2-EH^2}\)(pytago)=\(\sqrt{30^2-22,98^2}\)=19,29

mà:DF=DH+HF=13,28+19,29=32,57

chu vi \(_{\Delta DEF}\)=DE+EF+DF=26,54+30+32,57=89,11

\(S_{\Delta DEF}\)=\(\dfrac{EH.DF}{2}\)=\(\dfrac{22,98.32,57}{2}\)=374,2293