The altitudes of an acute-angled triangle $ABC$, drawn from vertices $A$ and $B$, intersect at point $H$, and $\angle AHB = 120^{\circ}$. The angle bisectors drawn from vertices $B$ and $C$ intersect at point $K$, and $\angle BKC = 130^{\circ}$. Find the angle $ABC$.
Since $\angle A H B=120^{\circ}$, and angle $A C B-$ is acute, then $\angle C=180^{\circ}-120^{\circ}=60^{\circ}$.

Since $\angle B K C=90^{\circ}+1 / 2 \angle A$, then $\angle A=80^{\circ}$.

Therefore, $\angle B=180^{\circ}-60^{\circ}-80^{\circ}=40^{\circ}$.

## Answer

$40^{\circ}$.