Bài 9:
a: Ta có: \(\hat{xOy}+\hat{yOz}=150^0\)
\(\hat{xOy}-\hat{yOz}=90^0\)
Do đó: \(\hat{xOy}=\frac{150^0+90^0}{2}=120^0;\hat{yOz}=120^0-90^0=30^0\)
b: Ta có: \(\hat{z^{\prime}Oy}+\hat{zOy}=180^0\) (hai góc kề bù)
=>\(\hat{z^{\prime}Oy}=180^0-30^0=150^0\)
=>\(\hat{z^{\prime}Oy}=\hat{xOz}\left(=150^0\right)\)
Bài 7:
Ta có: \(\hat{xOy}+\hat{xOm}=180^0\) (hai góc kề bù)
=>\(\hat{xOm}=180^0-65^0=115^0\)
Ta có: \(\hat{xOy}=\hat{mOn}\) (hai góc đối đỉnh)
mà \(\hat{xOy}=65^0\)
nên \(\hat{mOn}=65^0\)
Ta có: \(\hat{xOm}=\hat{yOn}\) (hai góc đối đỉnh)
mà \(\hat{xOm}=115^0\)
nên \(\hat{yOn}=115^0\)
Bài 6: Ta có: \(\hat{AOC}+\hat{BOD}=140^0\)
mà \(\hat{AOC}=\hat{BOD}\) (hai góc đối đỉnh)
nên \(\hat{AOC}=\hat{BOD}=\frac{140^0}{2}=70^0\)
Ta có: \(\hat{AOC}+\hat{AOD}=180^0\) (hai góc kề bù)
=>\(\hat{AOD}=180^0-70^0=110^0\)
Ta có: \(\hat{AOD}=\hat{BOC}\) (hai góc đối đỉnh)
mà \(\hat{AOD}=110^0\)
nên \(\hat{BOC}=110^0\)
