a)
b) 
c) Ta có: $\widehat{xOy}+\widehat{x'Oy}=180^{\circ}$ (hai góc kề bù )nên$\widehat{x'Oy}=180^{\circ}-\widehat{xOy}=180^{\circ}-90^{\circ}=90^{\circ}$
Ta lại có: $\widehat{x'Oy'}=\widehat{xOy};\widehat{xOy'}=\widehat{x'Oy}$ (các cặp góc đối đỉnh) nên $\widehat{x'Oy'}=90^{\circ}, \widehat{xOy'}=90^{\circ}$