a) Do hai góc xOy và yOz là hai góc kề nhau nên $\widehat{xOy}+\widehat{yOz}=\widehat{xOz}=150^{\circ}$

Mà $\widehat{xOy}−\widehat{yOz}=90^{\circ} nên\widehat{xOy}=(150^{\circ}+90^{\circ}):2=120^{\circ}$

=> $\widehat{yOz}150^{\circ}−\widehat{xOy}=150^{\circ}−120^{\circ}=30^{\circ}$

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: $\widehat{xOy′}=180^{\circ}−\widehat{xOy}=180^{\circ}−120^{\circ}=60^{\circ}$