Ta có: Ot là tia phân giác góc xOy nên ta có:

$\widehat{KOA}=\widehat{AOH}=\frac{\widehat{KOH}}{2}=\frac{{{80}^{0}}}{2}={{40}^{0}}$                                                    

Xét tam giác vuông KOA ta có:

$\widehat{KOA}+\widehat{OKA}+\widehat{KAO}={{180}^{0}}$

${{40}^{0}}+{{90}^{0}}+\widehat{KAO}={{180}^{0}}$

$\widehat{KAO}={{50}^{0}}$

Xét tam giác vuông HOA ta có:

$\widehat{HOA}+\widehat{OHA}+\widehat{HAO}={{180}^{0}}$

${{40}^{0}}+{{90}^{0}}+\widehat{HAO}={{180}^{0}}$

$\widehat{HAO}={{50}^{0}}$

Vậy $\widehat{KAH}+\widehat{HAO}={{100}^{0}}$