các bạn trả lời thì vẽ luôn hình nka

Xét \(\Delta ABC.có:\widehat{A}+\widehat{B}+\widehat{C}=180^o\Rightarrow\widehat{B}+\widehat{C}=105^o\)

\(\widehat{B}=\left(105^o+25^o\right):2=65^o\)

\(\widehat{C}=105^o-65^o=40^o\)

ta có \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\)

\(\Rightarrow75^o+\widehat{B}+\widehat{C}=180^o\\ \Rightarrow\widehat{B}+\widehat{C}=180^o-75^o=105^o\)

do đó:\(\widehat{B}=\dfrac{\left(105^o+25^o\right)}{2}=65^o\)

           \(\widehat{C}=65^o-25^o=40^o\)

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 \(\widehat{B}=\widehat{C}=55^0\)

Ta có: \(\widehat{ABC}=\widehat{ACB}\) ( cân tại đỉnh A )

\(\Rightarrow\left(\widehat{B}+30^0\right)+\widehat{B}+\widehat{B}=180^0\)

\(\Rightarrow3\widehat{B}+30^0=180^0\)

\(3\widehat{B}=180^0-30^0\)

\(3\widehat{B}=150^0\)

\(\Rightarrow\widehat{B}=50^0\)

\(\Rightarrow\widehat{C}=50^0\)

\(\Rightarrow\widehat{A}=50^0+30^0=80^0\)