Question

Cho tứ giác ABCD biết A\(\widehat{A}+\widehat{B}=220^o\), \(\widehat{B}+\widehat{C}=170^o,\widehat{A}+\widehat{C}=150^o\). Tính số đo các góc của rứ giác ABCD

Answer 1

Đặt góc A, góc B, góc C, góc D lần lượt là a,b,c,d

Theo đề, ta có: a+b=220; b+c=170; a+c=150 và a+b+c+d=360

\(\Leftrightarrow\left\{{}\begin{matrix}a=220-b\\c=170-b\\220-b+170-b=150\\a+b+c+d=360\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=220-b=100\\c=170-b=50\\b=120\\a+b+c+d=360\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=100\\b=120\\c=50\\d=90\end{matrix}\right.\)

Answer 2

Có: \(\widehat{A}+\widehat{B}=220^0,\widehat{B}+\widehat{C}=170^0,\widehat{A}+\widehat{C}=150^0\)

\(\Rightarrow2\left(\widehat{A}+\widehat{B}+\widehat{C}\right)=220^0+170^0+150^0\)

\(\Rightarrow\widehat{A}+\widehat{B}+\widehat{C}=270^0\)

Mà \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^0\)

\(\Rightarrow\widehat{D}=90^0\)

Có: \(\widehat{A}+\widehat{B}=220^0,\widehat{B}+\widehat{C}=170^0\)

\(\widehat{A}+\widehat{B}-\widehat{B}-\widehat{C}=220^0-170^0=50^0\)

\(\Rightarrow\widehat{A}-\widehat{C}=50^0\)

mà \(\widehat{A}+\widehat{C}=150^0\)

\(\Rightarrow\widehat{A}=100^0,\widehat{C}=50^0\)

mà \(\widehat{A}+\widehat{B}+\widehat{C}=270^0\)

\(\Rightarrow\widehat{B}=120^0\)