Lời giải:
Qua $K$ kẻ $Kt\parallel a\Rightarrow Kt\parallel b$
Vì $a\parallel Kt$ nên $\widehat{AKt}=\widehat{cAa}=40^0$ (hai góc đồng vị)
Vì $b\parallel Kt$ nên $\widehat{tKB}=\widehat{KBb'}=35^0$ (hai góc so le trong)
$\Rightarrow \widehat{AKB}=\widehat{AKt}+\widehat{tKB}=40^0+35^0=75^0$