Question

Cho xx' và yy' cắt nhau tại O. Tính góc x'Oy, góc xOy, x'Oy, góc xOy'.Biết:

a) xOy - x'Oy = 20^o

b) xOy + xOy=180^o

Answer 1

a: \(\left\{{}\begin{matrix}\widehat{xOy}-\widehat{x'Oy}=20^0\\\widehat{xOy}+\widehat{x'Oy}=180^0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\widehat{xOy}=\dfrac{20^0+180^0}{2}=100^0\\\widehat{x'Oy}=80^0\end{matrix}\right.\)

\(\widehat{x'Oy'}=\widehat{xOy}=100^0\)(hai góc đối đỉnh)

\(\widehat{x'Oy}=\widehat{xOy'}=80^0\)(hai góc đối đỉnh)

b: \(2\cdot\widehat{xOy}=180^0\)

nên \(\widehat{xOy}=90^0\)

=>\(\widehat{x'Oy}=\widehat{xOy'}=\widehat{x'Oy'}=90^0\)