Question

Tứ giác ABCD có góc B = góc A + \(10^o\), goc C = goc B + \(10^0\), góc D = goc C + \(10^0\). Tính góc B ?

Answer 1

Ta có : \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360\)

\(\Rightarrow\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=\widehat{A}+\widehat{A}+10^o+\widehat{B}+10^o+\widehat{C}+10^o\)

\(=\widehat{A}+\widehat{A}+10^o+\widehat{A}+10^o+10^o+\widehat{A}+10^o+10^o+10^o\)

\(\Rightarrow4\widehat{A}+60^o=360^o\)

\(\Rightarrow4\widehat{A}=360^o-60^o=300^o\)

\(\Rightarrow\widehat{A}=300^o:4=75^o\)

\(\Rightarrow\widehat{B}=\widehat{A}+10^o=75^o+10^o=85^o\)

Vậy....

Answer 2

tham khảo đây bn nè:

https://olm.vn/hoi-dap/tim-kiem?q=t%E1%BB%A9+gi%C3%A1c+ABCD+c%C3%B3+g%C3%B3c+B=A+10,+g%C3%B3c+C=B+10,+g%C3%B3c+D=C+10.+t%C3%ADnh+c%C3%A1c+g%C3%B3c&id=178966