Giải:
a) Vì góc A và góc B bù nhau
Nên \(\widehat{A}+\widehat{B}=180^0\)
\(\Leftrightarrow\widehat{A}=180^0-\widehat{B}\)
Lại có \(\widehat{A}-\widehat{B}=20^0\)
\(\Leftrightarrow180^0-\widehat{B}-\widehat{B}=20^0\)
\(\Leftrightarrow180^0-2\widehat{B}=20^0\)
\(\Leftrightarrow2\widehat{B}=160^0\)
\(\Leftrightarrow\widehat{B}=80^0\)
\(\Leftrightarrow\widehat{A}=180^0-\widehat{B}=180^0-80^0=100^0\)
Vậy ...
b) Vì tia Oz nằm giữa hai tia Ox và Oy
Nên ta có đẳng thức:
\(\widehat{xOz}+\widehat{yOz}=\widehat{xOy}\)
\(\Leftrightarrow\widehat{xOz}+\widehat{yOz}=90^0\) (*)
Lại có: \(4\widehat{xOz}=5\widehat{yOz}\)
\(\widehat{xOz}=\dfrac{5\widehat{yOz}}{4}=\dfrac{5}{4}\widehat{yOz}\)
Thay vào (*), ta được:
\(\Leftrightarrow\dfrac{5}{4}\widehat{yOz}+\widehat{yOz}=90^0\)
\(\Leftrightarrow\dfrac{9}{4}\widehat{yOz}=90^0\)
\(\Leftrightarrow\widehat{yOz}=40^0\)
\(\Leftrightarrow\widehat{xOz}=\widehat{xOy}-\widehat{yOz}=90^0-40^0=50^0\)
Vậy ...
