AD//BC
=>\(\widehat{A}+\widehat{B}=180^0\)
mà \(\widehat{A}-\widehat{B}=30^0\)
nên \(\widehat{A}=\dfrac{180^0+30^0}{2}=105^0;\widehat{B}=105^0-30^0=75^0\)
Ta có: AD//BC
=>\(\widehat{D}+\widehat{C}=180^0\)
=>\(\widehat{C}\cdot3+\widehat{C}=180^0\)
=>\(4\cdot\widehat{C}=180^0\)
=>\(\widehat{C}=45^0\)
=>\(\widehat{D}=3\cdot45^0=135^0\)