Do \(AB//CD\)
\( \Rightarrow \widehat A + \widehat D = {180^0}\) (hai góc trong cùng phía)
Ta có: \(\widehat A = 3\widehat D\) (gt)
\(\eqalign{
& \Rightarrow 3\widehat D + \widehat D = {180^0} \cr
& \Rightarrow \widehat D = {45^0} \cr
& \Rightarrow \widehat A = {3.45^0} = {135^0} \cr} \)
Do \(AB//CD\)
\(\Rightarrow \widehat B + \widehat C = {180^0}\) (hai góc trong cùng phía)
Mà: \(\widehat B - \widehat C = {30^0}\) (gt)
\(\eqalign{
& \Rightarrow 2\widehat B = {210^0} \Rightarrow \widehat B = {105^0} \cr
& \widehat C = \widehat B - {30^0} = {105^0} - {30^0} = {75^0} \cr} \)
