Ta có:
\(\widehat{A}+\widehat{B}+\widehat{C}=180^o\) (tổng 3 góc trong 1 tam giác)
\(\Rightarrow\left(180^o-3\widehat{C}\right)+\left(k\cdot\widehat{C}\right)+\widehat{C}=180^o\)
\(\Rightarrow180^o-3\widehat{C}+k\cdot\widehat{C}+\widehat{C}=180^o\)
\(\Rightarrow-3\widehat{C}+k\cdot\widehat{C}+\widehat{C}=180^o-180^o\)
\(\Rightarrow-3\widehat{C}+k\cdot\widehat{C}+\widehat{C}=0^o\)
\(\Rightarrow\left(-3+k+1\right)\cdot\widehat{C}=0^o\)
\(\Rightarrow\left(-2+k\right)\cdot\widehat{C}=0^o\)
\(\Rightarrow\left[{}\begin{matrix}-2+k=0^o\\\widehat{C}=0^o\end{matrix}\right.\)
Vì \(\widehat{C}\) là 1 góc của tam giác nên \(\widehat{C}\ne0^o\)
\(\Rightarrow-2+k=0^o\)
\(\Rightarrow k=0^o+2\)
\(\Rightarrow k=2^o\)
Ta có:
\(\widehat{A}+\widehat{B}+\widehat{C}=180^o\) (theo định lí tổng 3 góc trong tam giác)
\(\Rightarrow\widehat{A}=180^o-\left(\widehat{B}+\widehat{C}\right)\) \(\left(1\right)\)
mà \(\widehat{A}=180^o-3\widehat{C}\) (theo đề bài) \(\left(2\right)\)
\(\Rightarrow\widehat{B}+\widehat{C}=3\widehat{C}\Rightarrow\widehat{B}=3\widehat{C}-\widehat{C}\Rightarrow\widehat{B}=2\widehat{C}\)
mà \(\widehat{B}=k.\widehat{C}\) (theo đề bài)
\(\Rightarrow k=2\)
Vậy \(k=2\).
