Xét hình thang `ABCD`
Ta có:$\begin{cases}\dfrac{\hat{B}}{\hat{C}}=\dfrac{5}{4} \\\hat{B}+\hat{C}=180^o \end{cases}$
$⇔\begin{cases}4\hat{B}=5\hat{C} \\\hat{B}+\hat{C}=180^o \end{cases}$
$⇔\begin{cases}4(180^o-\hat{C})=5\hat{C} \\\hat{B}=180^o-\hat{C} \end{cases}$
$⇔\begin{cases}720^o-4\hat{C}=5\hat{C} \\\hat{B}=180^o-\hat{C} \end{cases}$
$⇔\begin{cases}720^o=9\hat{C} \\\hat{B}=180^o-\hat{C} \end{cases}$
$⇔\begin{cases}\hat{C}=80^o \\\hat{B}=180^o-80^o=100^o \end{cases}$
Ta lại có: $\begin{cases}\hat{A}=1,5\hat{D} \\\hat{A}+\hat{D}=180^o \end{cases}$
$⇔\begin{cases}180^o-\hat{D}=1,5\hat{D} \\\hat{A}=180^o-\hat{D} \end{cases}$
$⇔\begin{cases}180^o=2,5\hat{D} \\\hat{A}=180^o-\hat{D} \end{cases}$
$⇔\begin{cases}\hat{D}=72^o \\\hat{A}=180^o-72^o=108^o \end{cases}$
