Đáp án:
`\hat{A}=45^o, \hat{B}=115^o, \hat{C}=65^o, \hat{D}=135^o`
Giải:
Tứ giác `ABCD` có $AB \ // \ CD$ → $\begin{cases} \widehat{A}+\widehat{D}=180^o \\ \widehat{B}+\widehat{C}=180^o \end{cases}$
Mặt khác:
$\begin{cases} \widehat{A}=\dfrac{1}{3}\widehat{D} \\ \widehat{B}-\widehat{C}=50^o \end{cases} → \begin{cases} \dfrac{1}{3}\widehat{D}+\widehat{D}=180^o \\ \widehat{B}+\widehat{B}=180^o+50^o \end{cases}$
→ $\begin{cases} \dfrac{4}{3}\widehat{D}=180^o \\ 2\widehat{B}=230^o \end{cases} → \begin{cases} \widehat{D}=135^o \\ \widehat{B}=115^o \end{cases}$
→ $\begin{cases} \widehat{A}=\dfrac{1}{3}\widehat{D}=\dfrac{1}{3}.135^o=45^o \\ \widehat{C}=180^o-\widehat{B}=180^o-115^o=65^o \end{cases}$
