ta có:
$\widehat{ICD}+\widehat{IDC}+\widehat{CID}=180^{\circ}$
$=>\widehat{ICD}+\widehat{IDC}=\dfrac{\widehat{BCD}}{2}+\dfrac{\widehat{ADC}}{2}=180^{\circ}-105^{\circ}=75^{\circ}$
$=>\widehat{BCD}+\widehat{CDA}=150^{\circ}$
mà
$\widehat{A}+\widehat{B}+\widehat{BCD}+\widehat{CDA}=360^{\circ}$
$=>\widehat{A}=360^{\circ}-150^{\circ}-90^{\circ}$
$=>\widehat{A}=120^{\circ}$
ta có:
$\widehat{ICD}+\widehat{IDC}+\widehat{CID}=180^{\circ}$
$=>\widehat{ICD}+\widehat{IDC}=\dfrac{\widehat{BCD}}{2}+\dfrac{\widehat{ADC}}{2}=180^{\circ}-105^{\circ}=75^{\circ}$
$=>\widehat{BCD}+\widehat{CDA}=150^{\circ}$
mà
$\widehat{A}+\widehat{B}+\widehat{BCD}+\widehat{CDA}=360^{\circ}$
$=>\widehat{A}=360^{\circ}-150^{\circ}-90^{\circ}$
$=>\widehat{A}=120^{\circ}$