Ta có tổng 4 góc trong tứ giác là \(360^o\)
Hay: \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o\)
Mà: \(\widehat{D}=50^o;\widehat{B}=130^o;\widehat{A}=\dfrac{5}{4}\widehat{C}\)
Thay vào ta có:
\(\dfrac{5}{4}\widehat{C}+130^o+\widehat{C}+50^o=360^o\)
\(\Rightarrow\dfrac{9}{4}\widehat{C}+180^o=360^o\)
\(\Rightarrow\dfrac{9}{4}\widehat{C}=180^o\)
\(\Rightarrow\widehat{C}=180:\dfrac{9}{4}=80^o\)
Ta tìm được góc A:
\(\widehat{A}=\dfrac{5}{4}\widehat{C}=100^o\)
Theo định lý tổng 4 góc trong tứ giác :
\(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o\\ \Leftrightarrow\dfrac{5}{4}\widehat{C}+130^o+\widehat{C}+50^o=360^o\\ \Rightarrow\dfrac{9}{4}\widehat{C}=360^o-130^o-50^o\\ \Rightarrow\dfrac{9}{4}\widehat{C}=180^o\\ \Rightarrow\widehat{C}=80^o\)
\(\Rightarrow\widehat{A}=\dfrac{5}{4}\times80^o=100^o\)
