Ta có: \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\) (tổng 3 góc trong tam giác)
\(\Rightarrow90^o+35^o+\widehat{C}=180^o\)
\(\Rightarrow\widehat{C}=180^o-\left(90^o+35^o\right)=180^o-125^o=55^o\)
Vậy \(\widehat{C}=55^o\).
\(\Delta ABC\) có: \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\) (định lí)
\(\Leftrightarrow90^o+35^o+\widehat{C}=180^o\)
\(\Leftrightarrow125^o+\widehat{C}=180^o\)
\(\Leftrightarrow\widehat{C}=180^o-125^o=55^o\)
Vậy \(\widehat{C}=55^o.\)
Ta có: \(\widehat{C}+\widehat{B}=90^O\)( \(\Delta ABC\) vuông tại A)
\(\widehat{C}=90^0-35^0=55^0\)
Vậy \(\widehat{C}=55^0\)
