Xét ∆AKB ta có :
KAC + ACK + AKC = 180°
=> ACK = 40°
Mà IHK + IHC = 180° ( kề bù)
=> KHI = 130°
\(\widehat{KAI}+\widehat{AIH}+\widehat{IHK}+\widehat{HKA}=360^o\)
\(\widehat{IHK}=360^o-50^o-90^o-90^o\)
\(\Rightarrow\widehat{IHK}=130^o\)
\(\widehat{ACK}=180^o-50^0-90^o\)
\(\widehat{ACK}=40^o\)
\(\widehat{IHC}=180^o-90^o-40^o\)
\(\widehat{IHC}=50^o\)