$\square$ Answer: $65^{\circ}$.

Fig. 1. Let $\angle K L O=x$. Then $\angle K L A=2 x$ and $\angle L K A=50^{\circ}-2 x$. $K O$ is the bisector of the angle adjacent to $\angle L K A$, so $\angle A K O=90^{\circ}-\left(50^{\circ}-2 x\right) / 2=$ $65^{\circ}+x$. Then $\angle K O L=180^{\circ}-x-\left(50^{\circ}-2 x\right)-\left(65^{\circ}+x\right)=65^{\circ}$.
2. A circle with center $O$ is inscribed in triangle $A B C$. Point $L$ lies on the extension of side $A B$ beyond vertex $A$. A tangent line drawn from $L$ to the circle intersects side $A C$ at point $K$. Find $\angle K O L$, if $\angle B A C=50^{\circ}$.