Is the answer 120, 15, 45?

Please elaborate on how did you get the solution.

It's easy to prove that BCL is equilateral (use lemma - BL=IL=CL). Let AC, BH intersect at P, AC, AB - at Q. We have $ \angle QAC=\angle PAB=60^{\circ}$ => $ \angle HCP=\angle HPQ=30^{\circ}$ and $ \angle BHC=60^{\circ}$. Hence HCOB is inscribed quadrilateral, and HO is bisector of CHB. Now using AH=OH and $ \angle OHB=30^{\circ}$ we get $ \angle HBO=75^{\circ}$ => $ \angle ABC=15^{\circ}$ (because <HBQ= <CBO=30).

Angles $\angle LAB$ and $\angle LCB$ share the same arc, so $\angle LAB = \frac{1}{2} \angle C$. Similarly, $\angle LBA = \frac{1}{2} \angle C$. Using the same idea $\angle BLC = \angle A$ and $\angle ALC = \angle B$. Since $\angle LBI = \frac{1}{2} (\angle B + \angle C)$, then $\triangle LBI$ dictates that $\angle LIB = \angle LBI$ and so $IL=AL=BL=AB$. Or you may have already known this result. Since $\triangle ABL$ is equilateral, $\angle A + \angle B = 60^{\circ}$ and $\angle C = 120^{\circ}$. From observing the triangle formed by $A$, $B$, and the foot of $B$ onto $AH$, $\angle BAH = 90^{\circ} - \angle B$. Likewise, $\angle ABH = \angle 90^{\circ}-\angle A$, so $\angle AHB = \angle A + \angle B = 60^{\circ}$. From $AH=OH$, $\angle AOH = \angle OAH = 120^{\circ} - \angle B$, so $\angle AHO = 2\angle B - 60^{\circ}$. Since $\angle AOB = 2\angle ALB = 120^{\circ}$, $AOBH$ is cyclic, and so $\angle AHO = \angle ABO \Rightarrow 2 \angle B - 60^{\circ} = 30^{\circ} \Rightarrow \angle B = 45^{\circ}, \angle A = 15^{\circ}$.