ta có: \(\left\{{}\begin{matrix}\dfrac{A}{3}=\dfrac{B}{5}=\dfrac{C}{7}\\A+B+C=180^o\end{matrix}\right.\)
áp dụng t/c dtsbn ta có:
\(\dfrac{A}{3}=\dfrac{B}{5}=\dfrac{C}{7}=\dfrac{A+B+C}{3+5+7}=\dfrac{180^o}{15}=12^o\)
\(\dfrac{A}{3}=12^o\Rightarrow A=36^o\\ \dfrac{B}{5}=12^o\Rightarrow B=60^o\\ \dfrac{C}{7}=12^o\Rightarrow C=84^o\)
Áp dụng t/c dtsbn:
\(\dfrac{\widehat{A}}{3}=\dfrac{\widehat{B}}{5}=\dfrac{\widehat{C}}{7}=\dfrac{\widehat{A}+\widehat{B}+\widehat{C}}{3+5+7}=\dfrac{180^0}{15}=12^0\)
\(\Rightarrow\left\{{}\begin{matrix}\widehat{A}=12^0.3=36^0\\\widehat{B}=12^0.5=60^0\\\widehat{C}=12^0.7=84^0\end{matrix}\right.\)
