a) Do hai góc xOy và yOz là hai góc kề nhau nên $\widehat{xOy}+\widehat{yOz}=\widehat{xOz}=150^{\circ}$
Mà $\widehat{xOy}-\widehat{yOz}=90^{\circ}$ nên $\widehat{xOy}=(150^{\circ}+90^{\circ})/2=120^{\circ}$
Suy ra $\widehat{yOz}150^{\circ}-\widehat{xOy}=150^{\circ}-120^{\circ}=30^{\circ}$
b)Ta có $\widehat{x'Oy'}=\widehat{xOy}$ (hai góc đối đỉnh) nên $\widehat{x'Oy'}=120^{\circ}$
Ta có $\widehat{y'Oz}+\widehat{yOz}=180^{\circ}$(hai góc kề bù) nên $\widehat{y'Oz}=180-\widehat{yOz}=180^{\circ}-30^{\circ}=150^{\circ}$
Tương tự ta có: $\widehat{xOy'}=180^{\circ}-\widehat{xOy}=180^{\circ}-120^{\circ}=60^{\circ}$