( Hình em tự vẽ nhé )
+ Ta có: ΔABC = ΔDEF
=> \(\widehat{A}=\widehat{D}=30^o\)
+ Ta có: \(2\widehat{B}=3\widehat{C}\)
=> \(\widehat{B}=\dfrac{3\widehat{C}}{2}\)
+ Xét ΔABC
=> \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\left(t3g\Delta\right)\)
Mà \(\widehat{A}=30^o;\widehat{B}=\dfrac{3\widehat{C}}{2}\)
=> \(30^o+\dfrac{3\widehat{C}}{2}+\widehat{C}=180^o\)
=> \(\dfrac{3\widehat{C}}{2}+\widehat{C}=150^o\)
\(\Rightarrow\dfrac{3\widehat{C}}{2}+\dfrac{2\widehat{C}}{2}=150^o\)
\(\Rightarrow\dfrac{5\widehat{C}}{2}=150^o\)
\(\Rightarrow5\widehat{C}=75^o\)
\(\Rightarrow\widehat{C}=15^o\)
+ Xét ΔABC
\(\Rightarrow\widehat{A}+\widehat{B}+\widehat{C}=180^o\left(t3g\Delta\right)\)
\(\Rightarrow30^o+15^o+\widehat{B}=180^o\)
\(\Rightarrow\widehat{B}=135^o\)
