Đáp án:
$\begin{array}{l}
Do:\widehat A + \widehat B + \widehat C + \widehat D = {360^0}\\
Theo\,t.c\,day\,ti\,so = \,nhau\\
\widehat A:\widehat B:\widehat C:\widehat D = 4:3:2:1\\
\Rightarrow \dfrac{{\widehat A}}{4} = \dfrac{{\widehat B}}{3} = \dfrac{{\widehat C}}{2} = \dfrac{{\widehat D}}{1}\\
= \dfrac{{\widehat A + \widehat B + \widehat C + \widehat D}}{{4 + 3 + 2 + 1}} = \dfrac{{{{360}^0}}}{{10}} = {36^0}\\
\Rightarrow \left\{ \begin{array}{l}
\widehat A = {144^0}\\
\widehat B = {108^0}\\
\widehat C = {72^0}\\
\widehat D = {36^0}
\end{array} \right.
\end{array}$
