Question

cho △ABC có các góc A,B,C lần lượt với tỉ lệ 2,3,4 tính mỗi góc?

Answer 1

Áp dụng t.c dtsbn:

\(\dfrac{\widehat{A}}{2}=\dfrac{\widehat{B}}{3}=\dfrac{\widehat{C}}{4}=\dfrac{\widehat{A}+\widehat{B}+\widehat{C}}{2+3+4}=\dfrac{180^0}{9}=20^0\\ \Rightarrow\left\{{}\begin{matrix}\widehat{A}=40^0\\\widehat{B}=60^0\\\widehat{C}=80^0\end{matrix}\right.\)

Answer 2

Áp dụng t/c dtsbn:

\(\dfrac{\widehat{A}}{2}=\dfrac{\widehat{B}}{3}=\dfrac{\widehat{C}}{4}=\dfrac{\widehat{A}+\widehat{B}+\widehat{C}}{2+3+4}=\dfrac{180}{9}=20\)

\(\Rightarrow\left\{{}\begin{matrix}\widehat{A}=20^0.2=40^0\\\widehat{B}=20^0.3=60^0\\\widehat{C}=20^0.4=80^0\end{matrix}\right.\)