Question

Answer 1

\(\widehat{ABC}+\widehat{ABx}=180^0\)(hai góc kề bù)

=>\(\widehat{ABC}=180^0-110^0=70^0\)

Xét ΔABC có \(\widehat{A}=\widehat{C}\)

nên ΔABC cân tại B

=>\(\widehat{A}=\dfrac{180^0-\widehat{ABC}}{2}=\dfrac{180^0-70^0}{2}=55^0\)

Answer 2

\(\widehat{B}'=180-\widehat{B}=180^0-110^0=70^0\\ \widehat{A}=\widehat{C}=\left(180^0-70^0\right):2=55^0\)