Đáp án: $\,\widehat A = {165^0};\widehat B = {135^0};\widehat C = {45^0};\widehat D = {15^0}$
Giải thích các bước giải:
$\begin{array}{l}
Do:AB//CD\\
\Leftrightarrow \left\{ \begin{array}{l}
\widehat A + \widehat D = {180^0}\\
\widehat B + \widehat C = {180^0}\\
\widehat A + \widehat B + \widehat C + \widehat D = {360^0}
\end{array} \right.\\
Do:\widehat A = \widehat B + {30^0};\widehat C = 3\widehat D\\
\Leftrightarrow \widehat B + {30^0} + \widehat B + \widehat C + \dfrac{1}{3}\widehat C = {360^0}\\
\Leftrightarrow 2\widehat B + \dfrac{4}{3}\widehat C = {330^0}\\
\Leftrightarrow \left\{ \begin{array}{l}
\widehat B + \widehat C = {180^0}\\
2\widehat B + \dfrac{4}{3}\widehat C = {330^0}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\widehat B = {135^0}\\
\widehat C = {45^0}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\widehat A = \widehat B + {30^0} = {165^0}\\
\widehat D = \dfrac{1}{3}\widehat C = {15^0}
\end{array} \right.\\
Vậy\,\widehat A = {165^0};\widehat B = {135^0};\widehat C = {45^0};\widehat D = {15^0}
\end{array}$