Solution:
Given the circle;
[tex]\begin{gathered} mKH=188^o,mHI=50^o \\ \\ mKI=360^o-188^o-50^o \\ \\ mKI=122^o \\ \\ \text{ So from triangle OKI;} \\ \\ \angle K=\angle I \\ \\ \angle K=\frac{180^o-122^o}{2} \\ \\ \angle K=29^o \\ \\ m\angle LKI=90^o+29^o \\ \\ m\angle LKI=119^o \end{gathered}[/tex]
