Ta có: \(\Delta ABC\)cân \(\Rightarrow\widehat{B}=\widehat{C}\)
Xét \(\Delta ABC\)có:
\(\widehat{A}+\widehat{B}+\widehat{C}=180^o\)( đl tổng 3 góc của 1 tam giác)
hay \(70^o+\widehat{B}+\widehat{B}=180^o\)
\(\Rightarrow2\widehat{B}=180^o-70^o\)
\(\Rightarrow2\widehat{B}=110^o\)
\(\Rightarrow\widehat{B}=\frac{110^o}{2}=55^o\)
\(\Rightarrow\widehat{B}=\widehat{C}=55^o\)
✎✰ ๖ۣۜLαɗσηηα ༣✰✍ ghê :)) tớ ko nghĩ ra cách đs luôn :))
\(\Delta ABC:\widehat{A}+\widehat{B}+\widehat{C}\)
\(\Rightarrow\widehat{A}+\widehat{B}+\widehat{C}=180^0\)
Vì \(\Delta ABC\)cân tại A
Nên : \(\Rightarrow\widehat{B}=\widehat{C}=\frac{180^0-\widehat{A}}{2}=\frac{180^0-70^0}{2}=55^0\)
Vậy \(\widehat{B}=\widehat{C}=55^0\)
